bibkey: hanna2026a395833div authors: Paul D. Hanna year: 2026 title: “OEIS A395833, divisibility by the vanishing-diagonal exponent” doi: null url: https://oeis.org/A395833 claim: “G.f. A(x) satisfies [x^n] A( x/A(x)^(2n-1) ) = 0 for n > 1. Conjecture: (2n-1) divides a(n) for n >= 1.” strata_touched:

• D5/S1/Recurrence/Residue/DiagonalExponentSelfDivisibility license: citation-only triage: anchor

OEIS A395833: exponent divisibility

Paul D. Hanna’s entry specifies the generating equation and the divisibility conjecture quoted above. The normalized integer series has constant and linear coefficients one. Division by a power of this series means multiplication by its formal unit inverse.

For the family with exponent d*(n-1)+1, the coefficient of degree n is NegativePowerDiagonalModPrime.a (d+1) n: that construction uses parameter p with exponent (p-1)*(n-1)+1. Its generating equation and uniqueness identify the sequence, so A395833 is the parameter p=3, or slope d=2.

The general divisibility theorem follows by strong induction. For a smaller positive index m, the derivative coefficient identity gives e(n) dividing (n-m)*c, where c is the coefficient of degree n-m in A^(-e(n)*m). The affine identity e(m)=e(n)-d*(n-m) implies that e(n) divides e(m)*c. The induction hypothesis therefore makes each summand in the vanishing-diagonal recursion divisible by e(n). This proves the stated conjecture at slope two.

Verified locator

• URL: https://oeis.org/A395833
• The NAME and divisibility COMMENT are quoted in the claim above.