A173217 -id:A173217 - cp's OEIS frontend

A173219 G.f.: A(x) = Sum_{n>=0} (1 + x)^(n(n+1)/2) / 2^(n+1).

1, 2, 12, 124, 1800, 33648, 769336, 20796960, 648841680, 22945907520, 907036108432, 39631833652320, 1896696894062880, 98669609894805600, 5543804125505195040, 334563594743197602272, 21583554094995765302592

Offset: 0

Paul D. Hanna, Mar 05 2010

nonn

a(n) is the number of nonnegative integer matrices with n distinct columns and any number of nonzero rows with 2 ones in every column and columns in decreasing lexicographic order. - Andrew Howroyd, Jan 15 2020

Andrew Howroyd, Table of n, a(n) for n = 0..100

Row n=2 of A331278.

Cf. A121251, A173217, A173218.

Table[Sum[StirlingS1[n, j] * Sum[Binomial[j, s]*HurwitzLerchPhi[1/2, -j - s, 0], {s, 0, j}] / 2^(j+1), {j, 0, n}] / n!, {n, 0, 20}] (* Vaclav Kotesovec, Oct 08 2019 *)

{a(n)=local(A=sum(m=0,n^2+100,(1+x +O(x^(n+2)))^(m*(m+1)/2)/2^(m+1)));round(polcoeff(A,n))}

a(n) = A265937(n)/2. - Vaclav Kotesovec, Oct 08 2019
a(n) ~ 2^n * n^n / (2^(log(2)/4) * log(2)^(2*n+1) * exp(n)). - Vaclav Kotesovec, Oct 08 2019
a(n) = 2*A121251(n) for n > 0. - Andrew Howroyd, Jan 15 2020

A301310 G.f.: Sum_{n>=0} 2^n * (1+x)^(n^2) / 3^(n+1).

Original entry on oeis.org

1, 10, 360, 21840, 1857660, 203258160, 27188330400, 4298562686880, 784233322674120, 162161079972261480, 37477229047577953920, 9573364920705562944000, 2678416661190852872256960, 814535089079749159186189440, 267528376262254011309768677760, 94377360018309519999410315205120, 35590366640535756970223476489499280, 14287353028920891078189826021459809120

Offset: 0

Paul D. Hanna, Mar 18 2018

nonn

Is there a finite expression for the terms of this sequence?

a(n) is divisible by 10 for n>0 (conjecture).

			G.f.: A(x) = 1 + 10*x + 360*x^2 + 21840*x^3 + 1857660*x^4 + 203258160*x^5 + 27188330400*x^6 + 4298562686880*x^7 + 784233322674120*x^8 + ...
such that
A(x) = 1/3 + 2*(1+x)/3^2 + 2^2*(1+x)^4/3^3 + 2^3*(1+x)^9/3^4 + 2^4*(1+x)^16/3^5 + 2^5*(1+x)^25/3^6 + 2^6*(1+x)^36/3^7 + 2^7*(1+x)^49/3^8  + 2^8*(1+x)^64/3^9 + ...

Paul D. Hanna, Table of n, a(n) for n = 0..100

Cf. A173217, A301311.

Table[Sum[StirlingS1[n, j] * HurwitzLerchPhi[2/3, -2*j, 0]/3, {j, 0, n}] / n!, {n, 0, 20}] (* Vaclav Kotesovec, Mar 21 2018 *)

/* Continued fraction expression: */
{a(n) = my(CF=1, q = 1+x +x*O(x^n)); for(k=0, n, CF = 1/(3 - 2*q^(4*n-4*k+1)/(1 - 2*q^(2*n-2*k+1)*(q^(2*n-2*k+2) - 1)*CF)) ); polcoeff(CF, n)}
for(n=0, 30, print1(a(n), ", "))

G.f.: 1/(3 - 2*q/(1 - 2*q*(q^2-1)/(3 - 2*q^5/(1 - 2*q^3*(q^4-1)/(3 - 2*q^9/(1 - 2*q^5*(q^6-1)/(3 - 2*q^13/(1 - 2*q^7*(q^8-1)/(3 - ...))))))))) where q = (1+x), a continued fraction due to a partial elliptic theta function identity.
G.f.: Sum_{n>=0} 2^n/3^(n+1) * (1+x)^n * Product_{k=1..n} (3 - 2*(1+x)^(4*k-3)) / (3 - 2*(1+x)^(4*k-1)), due to a q-series identity.
a(n) = Sum_{k>=0} 2^k * binomial(k^2, n) / 3^(k+1).
a(n) ~ 2^(2*n + 1/2 + log(3/2)/8) * n^n / (3^(1 + log(3/2)/8) * exp(n) * (log(3/2))^(2*n + 1)). - Vaclav Kotesovec, Mar 21 2018

A265936 G.f.: Sum_{n>=0} (1 + x)^(n^2) / 2^n.

Original entry on oeis.org

2, 6, 72, 1488, 43212, 1615824, 73897824, 3995603040, 249332628600, 17635891224600, 1394325697514112, 121850733102557184, 11663364820483368384, 1213527023075625127296, 136368036713802512640384, 16459661773011642351224832, 2123742016843422531580031760, 291703805646180152870305600416, 42495460119330209128505618419584, 6544578588779477399509681497008256, 1062399800520315889891506552001161024, 181308080907736435566683700136306288320

Offset: 0

Paul D. Hanna, Dec 23 2015

nonn

			G.f.: A(x) = 2 + 6*x + 72*x^2 + 1488*x^3 + 43212*x^4 + 1615824*x^5 + 73897824*x^6 + 3995603040*x^7 + 249332628600*x^8 + 17635891224600*x^9 +...
where
A(x) = 1 + (1+x)/2 + (1+x)^4/2^2 + (1+x)^9/2^3 + (1+x)^16/2^4 + (1+x)^25/2^5 + (1+x)^36/2^6 + (1+x)^49/2^7 + (1+x)^64/2^8 +...+ (1+x)^(n^2)/2^n +...

Vaclav Kotesovec, Table of n, a(n) for n = 0..330

Cf. A265937, A303920, A173217.

Table[Round[Sum[Binomial[k^2, n]/2^k, {k, Sqrt[n], Infinity}]] , {n, 0, 20}] (* G. C. Greubel, May 23 2017 *)
Table[2*Sum[StirlingS1[n, j] * HurwitzLerchPhi[1/2, -2*j, 0]/2, {j, 0, n}] / n!, {n, 0, 20}] (* Vaclav Kotesovec, Oct 08 2019 *)

/* Informal listing of terms: */
{Vec( round( sum(n=0,600,(1+x +O(x^31))^(n^2)/2^n * 1.) ) )}
{Vec( round( sum(n=0,200, (1.+x)^n/2^n * prod(k=1,n, (2 - (1+x)^(4*k-3)) / (2 - (1+x)^(4*k-1)) +O(x^21) ) ) ) )}

G.f.: Sum_{n>=0} (1+x)^n/2^n * Product_{k=1..n} (2 - (1+x)^(4*k-3)) / (2 - (1+x)^(4*k-1)) due to a q-series identity.
G.f.: 1/(1 - (1+x)/2 /(1 - (1+x)*((1+x)^2-1)/2 /(1 - (1+x)^5/2 /(1 - (1+x)^3*((1+x)^4-1)/2 /(1 - (1+x)^9/2 /(1 - (1+x)^5*((1+x)^6-1)/2 /(1 - (1+x)^13/2 /(1 - (1+x)^7*((1+x)^8-1)/2 /(1 - ...))))))))), a continued fraction due to a partial elliptic theta function identity.
a(n) = Sum_{k>=sqrt(n)} binomial(k^2,n) / 2^k.
a(n) = Sum_{k=0..2*n} A303920(n,k) * 2^k, for n>0.
a(n) = 2 * A173217(n) for n>=0.
a(n) ~ 2^(2*n + 1/2 - log(2)/8) * n^n / (exp(n) * log(2)^(2*n + 1)). - Vaclav Kotesovec, Oct 08 2019

A300279 G.f.: Sum_{n>=0} (1 + x*(1+x)^n)^n / 2^(n+1).

Original entry on oeis.org

1, 1, 4, 16, 86, 544, 3904, 31328, 276798, 2660564, 27576614, 306051500, 3615559236, 45241980928, 597141146374, 8283583741588, 120393776421550, 1828261719906800, 28937578248560784, 476355010859517352, 8139464481630136242, 144109168217154747856, 2639508261422244889106, 49940898467864797567140, 974790619672853340925800

Offset: 0

Paul D. Hanna, Mar 01 2018

nonn

Row sums of triangle A300280.

			G.f.: A(x) = 1 + x + 4*x^2 + 16*x^3 + 86*x^4 + 544*x^5 + 3904*x^6 + 31328*x^7 + 276798*x^8 + 2660564*x^9 + 27576614*x^10 + ...
such that
A(x) = 1/2 + (1 + x*(1+x))/2^2 + (1 + x*(1+x)^2)^2/2^3 + (1 + x*(1+x)^3)^3/2^4 + (1 + x*(1+x)^4)^4/2^5 + (1 + x*(1+x)^5)^5/2^6 + (1 + x*(1+x)^6)^6/2^7 + ...
Also, due to a series identity,
A(x) = 1 + x*(1+x)/(2 - (1+x))^2 + x^2*(1+x)^4/(2 - (1+x)^2)^3 + x^3*(1+x)^9/(2 - (1+x)^3)^4 + x^4*(1+x)^16/(2 - (1+x)^4)^5 + x^5*(1+x)^25/(2 - (1+x)^5)^6 + x^6*(1+x)^36/(2 - (1+x)^6)^7 + ... + x^n * (1+x)^(n^2) / (2 - (1+x)^n)^(n+1) + ...
Triangle A300280 is defined by
T(n,k) = Sum_{j>=0} C(j+k, k) * C((j+k)*k, n-k) / 2^(j+k+1), begins:
1;
0, 1;
0, 3, 1;
0, 5, 10, 1;
0, 7, 57, 21, 1;
0, 9, 252, 246, 36, 1;
0, 11, 969, 2158, 710, 55, 1;
0, 13, 3414, 15927, 10260, 1635, 78, 1; ...
the row sums of which form this sequence.
RELATED INFINITE SERIES.
At x = -1/2: the following sums are equal
S1 = Sum_{n>=1} (2^n - 1)^(n-1) / 2^(n^2),
S1 = Sum_{n>=1} (-1)^(n-1) / (2^n - 1)^n.
Explicitly,
S1 = 1/2 + 3/2^4 + 7^2/2^9 + 15^3/2^16 + 31^4/2^25 + 63^5/2^36 + 127^6/2^49 + 255^7/2^64 + 511^8/2^81 + 1023^9/2^100 + 2047^10/2^121 + 4095^11/2^144 + ...
S1 = 1 - 1/3^2 + 1/7^3 - 1/15^4 + 1/31^5 - 1/63^6 + 1/127^7 - 1/255^8 + 1/511^9 - 1/1023^10 + 1/2047^11 - 1/4095^12 + 1/8191^13 - 1/16383^14 + ...
where S1 = 0.891784622610953349715890136060239421022216970366139189336822360...

Paul D. Hanna, Table of n, a(n) for n = 0..300

Cf. A302765, A300280, A173217, A300050.

{a(n) = my(A = sum(m=0, n, x^m * (1+x)^(m^2) / (2 - (1 + x + x*O(x^n))^m )^(m+1) )); polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))

G.f. is given by:
(1) Sum_{n>=0} (1 + x*(1+x)^n)^n / 2^(n+1).
(2) Sum_{n>=0} x^n * (1+x)^(n^2) / (2 - (1+x)^n)^(n+1).
Formulas for terms.
a(n) = Sum_{k=0..n} Sum_{j>=0} C(j+k, k) * C((j+k)*k, n-k) / 2^(j+k+1).

A303920 G.f.: A(x,y) = (1-y) * Sum_{n>=0} y^n * (1 + x*(1-y)^2)^(n^2).

Original entry on oeis.org

1, 0, 1, 1, 0, 0, 6, 6, 0, 0, 0, 4, 56, 56, 4, 0, 0, 0, 1, 117, 722, 722, 117, 1, 0, 0, 0, 0, 126, 2982, 12012, 12012, 2982, 126, 0, 0, 0, 0, 0, 84, 6916, 79548, 246092, 246092, 79548, 6916, 84, 0, 0, 0, 0, 0, 36, 10900, 312880, 2322000, 6002824, 6002824, 2322000, 312880, 10900, 36, 0, 0, 0, 0, 0, 9, 12717, 864009, 13617765, 74916306, 170048394, 170048394, 74916306, 13617765, 864009, 12717, 9, 0, 0, 0, 0, 0, 1, 11421, 1825786, 57282026, 604000555, 2669115383, 5489377628, 5489377628, 2669115383, 604000555, 57282026, 1825786, 11421, 1, 0, 0

Offset: 0

Paul D. Hanna, May 02 2018

G.f. A(x,y) = Sum_{n>=0} Sum_{k=0..2*n} T(n,k) * x^n*y^k, where T(n,k) is the term of this triangle at position k in row n.

			G.f.: A(x,y) = (1-y) * ( 1 + y*(1 + x*(1-y)^2) + y^2*(1 + x*(1-y)^2)^4 + y^3*(1 + x*(1-y)^2)^9 + y^4*(1 + x*(1-y)^2)^16 + y^5*(1 + x*(1-y)^2)^25 + ... ).
Explicitly,
A(x,y) = 1 + x*(y + y^2) + x^2*(6*y^2 + 6*y^3) + x^3*(4*y^2 + 56*y^3 + 56*y^4 + 4*y^5) + x^4*(y^2 + 117*y^3 + 722*y^4 + 722*y^5 + 117*y^6 + y^7) + x^5*(126*y^3 + 2982*y^4 + 12012*y^5 + 12012*y^6 + 2982*y^7 + 126*y^8) + x^6*(84*y^3 + 6916*y^4 + 79548*y^5 + 246092*y^6 + 246092*y^7 + 79548*y^8 + 6916*y^9 + 84*y^10) + x^7*(36*y^3 + 10900*y^4 + 312880*y^5 + 2322000*y^6 + 6002824*y^7 + 6002824*y^8 + 2322000*y^9 + 312880*y^10 + 10900*y^11 + 36*y^12) + x^8*(9*y^3 + 12717*y^4 + 864009*y^5 + 13617765*y^6 + 74916306*y^7 + 170048394*y^8 + 170048394*y^9 + 74916306*y^10 + 13617765*y^11 + 864009*y^12 + 12717*y^13 + 9*y^14) + x^9*(y^3 + 11421*y^4 + 1825786*y^5 + 57282026*y^6 + 604000555*y^7 + 2669115383*y^8 + 5489377628*y^9 + 5489377628*y^10 + 2669115383*y^11 + 604000555*y^12 + 57282026*y^13 + 1825786*y^14 + 11421*y^15 + y^16) + ...
This triangle begins:
[1];
[0, 1, 1];
[0, 0, 6, 6, 0];
[0, 0, 4, 56, 56, 4, 0];
[0, 0, 1, 117, 722, 722, 117, 1, 0];
[0, 0, 0, 126, 2982, 12012, 12012, 2982, 126, 0, 0];
[0, 0, 0, 84, 6916, 79548, 246092, 246092, 79548, 6916, 84, 0, 0];
[0, 0, 0, 36, 10900, 312880, 2322000, 6002824, 6002824, 2322000, 312880, 10900, 36, 0, 0];
[0, 0, 0, 9, 12717, 864009, 13617765, 74916306, 170048394, 170048394, 74916306, 13617765, 864009, 12717, 9, 0, 0];
[0, 0, 0, 1, 11421, 1825786, 57282026, 604000555, 2669115383, 5489377628, 5489377628, 2669115383, 604000555, 57282026, 1825786, 11421, 1, 0, 0]; ...

Paul D. Hanna, Table of n, a(n) for n = 0..2600, of rows 0..50, as a flattened triangle read by rows.

Cf. A303921 (diagonal), A303922 (column sums), A001813 (row sums), A265936 (y=2), A173217.

/* G.f. by Definition: */
{T(n,k) = my(A = (1-y) * sum(m=0,2*n, y^m * (1 + x*(1-y)^2  +x*O(x^(2*n)) )^(m^2))); polcoeff(polcoeff(A, n,x),k,y)}
for(n=0, 10, for(k=0,2*n, print1(T(n,k), ", ")); print(""))

/* Continued fraction expression: */
{T(n,k) = my(CF=1, q = 1 + x*(1-y)^2 +x*O(x^(2*n))); for(k=0, n, CF = 1/(1 - q^(4*n-4*k+1)*y/(1 - q^(2*n-2*k+1)*(q^(2*n-2*k+2) - 1)*y*CF)) ); polcoeff(polcoeff((1-y)*CF, n,x),k,y)}
for(n=0, 10, for(k=0,2*n, print1(T(n,k), ", ")); print(""))

/* G.f. by q-series identity: */
{T(n,k) = my(A =1, q = 1 + x*(1-y)^2 +x*O(x^(2*n))); A = (1-y) * sum(m=0,2*n, y^m*q^m * prod(k=1,m, (1 - y*q^(4*k-3)) / (1 - y*q^(4*k-1) +x*O(x^(2*n))) )); polcoeff(polcoeff(A, n,x),k,y)}
for(n=0, 10, for(k=0,2*n, print1(T(n,k), ", ")); print(""))

GENERATING FUNCTIONS.
(1) A(x,y) = (1-y) * Sum_{n>=0} y^n * (1 + x*(1-y)^2)^(n^2).
(2) A(x,y) = (1-y) * Sum_{n>=0} y^n * q^n * Product_{k=1..n} (1 - q^(4*k-3)*y) / (1 - q^(4*k-1)*y), where q = 1 + x*(1-y)^2, due to a q-series identity.
(3) A(x,y) = (1-y)/(1 - q*y/(1 - q*(q^2-1)*y/(1 - q^5*y/(1 - q^3*(q^4-1)*y/(1 - q^9*y/(1- q^5*(q^6-1)*y/(1 - q^13*y/(1 - q^7*(q^8-1)*y/(1 - ...))))))))), where q = 1 + x*(1-y)^2, a continued fraction due to an identity of a partial elliptic theta function.
FORMULAS INVOLVING TERMS.
Sum_{k=0..2*n} T(n,k)  =  (2*n)!/n!, for n>=0 (row sums = A001813).
Sum_{k=0..2*n} T(n,k) * (-1)^k  =  0, for n>=1 (symmetric rows).
Sum_{k=0..2*n} T(n,k) * 2^k  =  A265936(n), for n>=1.
Sum_{k=0..2*n} T(n,k) / 2^k  =  A173217(n) / 4^n, for n>=0.
Sum_{j=0..k^2} T(j,k) = A303922(k), for k>=0 (column sums).
T(n,n) = A303921(n), for n>=0 (diagonal).

A173218 G.f.: A(x) = Sum_{n>=0} (1 + x)^(n^2+n) / 2^(n+1).

Original entry on oeis.org

1, 4, 50, 1040, 30300, 1135080, 51972668, 2812429632, 175606496520, 12426817517920, 982846762742416, 85916923493646752, 8225856593959648696, 856044724445883011520, 96213518828394481754400

Offset: 0

Paul D. Hanna, Mar 05 2010

nonn

Vaclav Kotesovec, Table of n, a(n) for n = 0..300

Cf. A173217, A173219.

Table[Sum[StirlingS1[n, j] * Sum[Binomial[j, s] * HurwitzLerchPhi[1/2, -j - s, 0], {s, 0, j}], {j, 0, n}] / (2*n!), {n, 0, 20}] (* Vaclav Kotesovec, Oct 08 2019 *)

{a(n)=local(A=sum(m=0,n^2+100,(1+x +O(x^(n+2)))^(m*(m+1))/2^(m+1)));round(polcoeff(A,n))}

a(n) ~ 2^(2*n) * n^n / (2^(log(2)/8) * log(2)^(2*n+1) * exp(n)). - Vaclav Kotesovec, Oct 08 2019

A301466 a(n) = Sum_{k>=0} binomial(k^3, n)/2^(k+1).

Original entry on oeis.org

1, 13, 2335, 1178873, 1168712311, 1916687692685, 4697337224419543, 16082097033630615185, 73313708225823014181097, 429319086610079876821621425, 3140585308524019620784003889263, 28066697522114849327295724261347841, 300886927215791917153044786581553617063

Offset: 0

Vaclav Kotesovec, Mar 21 2018

nonn

Vaclav Kotesovec, Table of n, a(n) for n = 0..175

Cf. A173217, A301310, A301432, A301468.

Table[Sum[Binomial[k^3, n]/2^(k+1), {k, 0, Infinity}], {n, 0, 15}]
Table[Sum[StirlingS1[n, j] * HurwitzLerchPhi[1/2, -3*j, 0]/2, {j, 0, n}] / n!, {n, 0, 15}]

a(n) ~ 3^(3*n + 1/2) * n^(2*n) / (2 * exp(2*n) * (log(2))^(3*n + 1)).

G.f.: Sum_{n>=0} (1 + x)^(n^3) / 2^(n+1).

A301432 G.f.: Sum_{n>=0} 3^n * (1+x)^(n^2) / 4^(n+1).

Original entry on oeis.org

1, 21, 1512, 182448, 30845052, 6706403424, 1782361433664, 559861341721920, 202922346528231120, 83358099246202940880, 38271708686845732234752, 19421327571536329073316864, 10794249397953336851774993664, 6521104275997643157262604783616, 4254768377324045826054766465227264, 2981719456871640091643441908508931072

Offset: 0

Paul D. Hanna, Mar 21 2018

nonn

a(n) is divisible by 21 for n>0 (conjecture).

In general, for s > 1, Sum_{k>=0} binomial(k^2, n) / s^k is asymptotic to 2^(2*n + 1/2) * n^n / (s^(log(s)/8) * exp(n) * (log(s))^(2*n + 1)). - Vaclav Kotesovec, Mar 21 2018

			G.f.: A(x) = 1 + 21*x + 1512*x^2 + 182448*x^3 + 30845052*x^4 + 6706403424*x^5 + 1782361433664*x^6 + 559861341721920*x^7 + ...
such that
A(x) = 1/4 + 3*(1+x)/4^2 + 3^2*(1+x)^4/4^3 + 3^3*(1+x)^9/4^4 + 3^4*(1+x)^16/4^5 + 3^5*(1+x)^25/4^6 + 3^6*(1+x)^36/4^7 + 3^7*(1+x)^49/4^8  + 3^8*(1+x)^64/4^9 + ...

Paul D. Hanna, Table of n, a(n) for n = 0..100

Cf. A301310, A173217, A301311.

Table[Sum[StirlingS1[n, j] * HurwitzLerchPhi[3/4, -2*j, 0]/4, {j, 0, n}] / n!, {n, 0, 20}] (* Vaclav Kotesovec, Mar 21 2018 *)

/* Continued fraction expression: */
{a(n) = my(CF=1, q = 1+x +x*O(x^n)); for(k=0, n, CF = 1/(4 - 3*q^(4*n-4*k+1)/(1 - 3*q^(2*n-2*k+1)*(q^(2*n-2*k+2) - 1)*CF)) ); polcoeff(CF, n)}
for(n=0, 30, print1(a(n), ", "))

G.f.: 1/(4 - 3*q/(1 - 3*q*(q^2-1)/(4 - 3*q^5/(1 - 3*q^3*(q^4-1)/(4 - 3*q^9/(1 - 3*q^5*(q^6-1)/(4 - 3*q^13/(1 - 3*q^7*(q^8-1)/(4 - ...))))))))) where q = (1+x), a continued fraction due to a partial elliptic theta function identity.
G.f.: Sum_{n>=0} 3^n/4^(n+1) * (1+x)^n * Product_{k=1..n} (4 - 3*(1+x)^(4*k-3)) / (4 - 3*(1+x)^(4*k-1)), due to a q-series identity.
a(n) = Sum_{k>=0} 3^k * binomial(k^2, n) / 4^(k+1).
a(n) ~ 2^(2*n - 3/2) * n^n / ((4/3)^(log(4/3)/8) * exp(n) * (log(4/3))^(2*n + 1)). - Vaclav Kotesovec, Mar 21 2018

A301468 a(n) = Sum_{k>=0} binomial(k^4, n)/2^(k+1).

Original entry on oeis.org

1, 75, 272880, 4681655040, 221478589107480, 22313622005672849712, 4108665216956980742226192, 1249503956658157724969373808320, 583952821303314451291898006535866460, 397372225886096887788939487944785734626120, 377577476850495509525002042506806447493291890064

Offset: 0

Vaclav Kotesovec, Mar 21 2018

nonn

In general, for m > 2, Sum_{k>=0} binomial(k^m, n) / 2^(k+1) is asymptotic to m^(m*n + 1/2) * n^((m-1)*n) / (2*exp((m-1)*n) * (log(2))^(m*n + 1)).

Cf. A173217 (m=2), A301466 (m=3), A301310.

Table[Sum[Binomial[k^4, n]/2^(k+1), {k, 0, Infinity}], {n, 0, 12}]
Table[Sum[StirlingS1[n, j] * HurwitzLerchPhi[1/2, -4*j, 0]/2, {j, 0, n}] / n!, {n, 0, 12}]

a(n) ~ 2^(8*n) * n^(3*n) / (exp(3*n) * (log(2))^(4*n+1)).

A304860 G.f. A(x) satisfies: x = Sum_{n>=0} ( (1+x)^(n^2) - A(x)^n ) / 2^(n+1).

Original entry on oeis.org

1, 2, 32, 608, 17750, 683504, 32183336, 1782735248, 113381031512, 8138225237204, 650735042088080, 57369033007665680, 5529284312514428840, 578479328396134930928, 65297339893598788494368, 7910610591246432715704704, 1023854667471171305890388408, 141001918216059025744295715872, 20587944237516075824024078357264, 3176963079503660078673757802123360

Offset: 0

Paul D. Hanna, May 28 2018

nonn

			G.f.: A(x) = 1 + 2*x + 32*x^2 + 608*x^3 + 17750*x^4 + 683504*x^5 + 32183336*x^6 + 1782735248*x^7 + 113381031512*x^8 + 8138225237204*x^9 + ...
such that
x = ((1+x) - A(x))/2^2 + ((1+x)^4 - A(x)^2)/2^3 + ((1+x)^9 - A(x)^3)/2^4 + ((1+x)^16 - A(x)^4)/2^5 + ((1+x)^25 - A(x)^5)/2^6 + ((1+x)^36 - A(x)^6)/2^7 + ...
RELATED SERIES.
G(x) = Sum_{n>=0} (1+x)^(n^2) / 2^(n+1) = 1 + 3*x + 36*x^2 + 744*x^3 + 21606*x^4 + 807912*x^5 + 36948912*x^6 + 1997801520*x^7 + 124666314300*x^8 + ... + A173217(n)*x^n + ...
1/(2 - A(x)) = G(x) - x = 1 + 2*x + 36*x^2 + 744*x^3 + 21606*x^4 + 807912*x^5 + 36948912*x^6 + 1997801520*x^7 + 124666314300*x^8 + ...
Let F(x) satisfy
x = Sum_{n>=0} ( F(x)^n - A(x)^n ) / 2^(n+1), then
F(x) = 1 + 3*x + 27*x^2 + 555*x^3 + 16737*x^4 + 652815*x^5 + 30967917*x^6 + 1724292411*x^7 + 110091861729*x^8 + 7926482395935*x^9 + ...
where 1/(2 - F(x)) = x + 1/(2 - A(x)).

Paul D. Hanna, Table of n, a(n) for n = 0..60

Cf. A173217.

G.f. A(x) satisfies:
(1) x = Sum_{n>=0} ( (1+x)^(n^2) - A(x)^n ) / 2^(n+1).
(2) A(x) = 2 - 1/(G(x) - x), where G(x) = Sum_{n>=0} (1+x)^(n^2) / 2^(n+1) is the g.f. of A173217.

cp's OEIS Frontend

A173219 G.f.: A(x) = Sum_{n>=0} (1 + x)^(n(n+1)/2) / 2^(n+1).

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A301310 G.f.: Sum_{n>=0} 2^n * (1+x)^(n^2) / 3^(n+1).

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A265936 G.f.: Sum_{n>=0} (1 + x)^(n^2) / 2^n.

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A300279 G.f.: Sum_{n>=0} (1 + x*(1+x)^n)^n / 2^(n+1).

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A303920 G.f.: A(x,y) = (1-y) * Sum_{n>=0} y^n * (1 + x*(1-y)^2)^(n^2).

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A173218 G.f.: A(x) = Sum_{n>=0} (1 + x)^(n^2+n) / 2^(n+1).

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A301466 a(n) = Sum_{k>=0} binomial(k^3, n)/2^(k+1).

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A301432 G.f.: Sum_{n>=0} 3^n * (1+x)^(n^2) / 4^(n+1).

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