A391910 - OEIS

%I #15 Jan 17 2026 16:33:17

%S 1,1,2,1,8,16,1,26,192,384,1,80,1696,12288,24576,1,242,13440,272640,

%T 1966080,3932160,1,728,101296,5222400,104816640,754974720,1509949440,

%U 1,2186,743232,92663424,4693032960,93943234560,676457349120,1352914698240,1,6560,5367616,1570480128,190752178176,9618377932800,192409837240320,1385384650997760,2770769301995520

%N Triangle read by rows: T(n,k) are the unique integer coefficients such that Sum_{k=0..n} T(n,k)*A125790(k,m)/2^(n*k) = (m+2)^n for all n >= 0, m >= 0.

%C Equivalently, T(n,k) are the unique integer coefficients such that Sum_{k=0..n} T(n,k)*A125790(k,m)*(-1)^(n+k)/2^(n*k) = m^n for all n >= 0, m >= 0. - _Mikhail Kurkov_, Jan 12 2026

%H MathOverflow, <a href="https://mathoverflow.net/a/505968/580209">Recurrence for columns of A125790</a>, 2025.

%H MathOverflow, <a href="https://mathoverflow.net/a/504889/580209">Proof of a recurrent formula if the closed form is known</a>, 2025.

%H MathOverflow, <a href="https://mathoverflow.net/a/506732/580209">Pair of identities for A125790</a>, 2026.

%F T(n,k) = Sum_{j=0..2^k-1} (2*j+1)^n*(-1)^(A000120(j)+k).

%F T(n,k) = Sum_{j=0..n-1} binomial(n,j)*2^(k*(n-j))*T(j,k-1) for k > 0 with T(n,0) = 1.

%F T(n,k) = 2*Sum_{j=0..floor((n-k)/2)} binomial(n,k+2*j)*2^(k*(n-k-2*j))*T(k+2*j,k-1) for k > 0 with T(n,0) = 1.

%F E.g.f. for column k: exp(x)*Product_{j=1..k} (exp(2^j*x) - 1).

%e Triangle begins:

%e   1;

%e   1,    2;

%e   1,    8,     16;

%e   1,   26,    192,      384;

%e   1,   80,   1696,    12288,      24576;

%e   1,  242,  13440,   272640,    1966080,     3932160;

%e   1,  728, 101296,  5222400,  104816640,   754974720,   1509949440;

%e   1, 2186, 743232, 92663424, 4693032960, 93943234560, 676457349120, 1352914698240;

%t T[n_,k_]:=Sum[(2j+1)^n*(-1)^(DigitCount[j,2,1]+k),{j,0,2^k-1}];Table[T[n,k],{n,0,8},{k,0,n}]//Flatten (* _James C. McMahon_, Dec 30 2025 *)

%o (PARI) rows(n) = my(v1 = vector(n+1, i, vector(i, j, 0)), v2 = v1); v1[1][1] = 1; v2[1][1] = 1; for(i=1, n, v1[i+1][1] = 1; v2[i+1][1] = 1; v2[i+1][i+1] = 1); for(i=2, n, for(j=1, i-1, v2[i+1][j+1] = v2[i][j+1] + v2[i][j])); for(i=1, n, for(j=i, n, v1[j+1][i+1] = 2*sum(k=0, (j-i)\2, v2[j+1][i+2*k+1]*v1[i+2*k+1][i]*2^(i*(j-i-2*k))))); v1

%Y Cf. A000120, A125790.

%K nonn,tabl

%O 0,3

%A _Mikhail Kurkov_, Dec 23 2025