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bibkey: barry2005a105595 authors: Paul Barry year: 2005 title: “OEIS A105595, Row sums of number triangle A105594” doi: null url: https://oeis.org/A105595 claim: “%N Row sums of number triangle A105594. %C Conjecture : all terms are odd. %F a(n)=sum{k=0..n, mod(sum{j=0..n, abs(mu(binomial(n, j)))*mod(binomial(j, k), 2)}, 2)} %F A105594 T(n, k) = mod(Sum_{j=0..n}(abs(mu(binomial(n,j)))*mod(binomial(j,k),2)), 2).” strata_touched:

  • D5/S3/ArithSums/BarryTriangleRowSumsOdd license: citation-only triage: anchor

OEIS A105595 and A105594

OEIS A105595 records the NAME (%N, verbatim):

Row sums of number triangle A105594.

Its COMMENT conjecture (%C, verbatim) is:

Conjecture : all terms are odd.

Its FORMULA (%F, verbatim) is:

a(n)=sum{k=0..n, mod(sum{j=0..n, abs(mu(binomial(n, j)))*mod(binomial(j, k), 2)}, 2)}

OEIS A105594 records the FORMULA (%F, verbatim):

T(n, k) = mod(Sum_{j=0..n}(abs(mu(binomial(n,j)))*mod(binomial(j,k),2)), 2).

The AUTHOR line is _Paul Barry_, Apr 14 2005. The conjecture comment is unsigned, so it is attributed here to Paul Barry, the author of the entry and its NAME.

The theorem result settles the conjecture for the two displayed %F sums. The proof works modulo two. Its delicate step is that the outer sum contains the reduction modulo two of each inner sum before summation: each reduced value is first cast into ZMod 2, where x mod 2 and x agree, and only then are the finite sums interchanged. The Pascal row sum is 2^j; terms with k > j vanish, and modulo two only j=0 survives, with abs(mu(binomial(n,0))) = abs(mu(1)) = 1.

The %N text for A105594 also describes a matrix product, abs(A103447)*A047999 mod 2. No equivalence between that phrasing and the displayed %F sum is claimed. No other property of A105594 is settled here. The bounded scan for n=0 through n=120 found all 121 row sums odd, with maximum row sum 47; this finite calculation carries no proof of the universal statement.

Verified locator

  • URL: https://oeis.org/A105595
  • A105595 NAME (%N, verbatim): Row sums of number triangle A105594.
  • A105595 COMMENT (%C, verbatim): Conjecture : all terms are odd.
  • A105595 FORMULA (%F, verbatim): a(n)=sum{k=0..n, mod(sum{j=0..n, abs(mu(binomial(n, j)))*mod(binomial(j, k), 2)}, 2)}
  • A105594 FORMULA (%F, verbatim): T(n, k) = mod(Sum_{j=0..n}(abs(mu(binomial(n,j)))*mod(binomial(j,k),2)), 2).
  • AUTHOR (%A, verbatim): Paul Barry, Apr 14 2005