A136516
a(n) = (2^n+1)^n.
Original entry on oeis.org
1, 3, 25, 729, 83521, 39135393, 75418890625, 594467302491009, 19031147999601100801, 2460686496619787545743873, 1280084544196357822418212890625, 2672769719437237714909813214827010049, 22366167213460480200139104627873703828439041
Offset: 0
A(x) = 1 + 3x + 5^2*x^2/2! + 9^3*x^3/3! + 17^4*x^4/4! +... + (2^n+1)^n*x^n/n! +...
A(x) = exp(x) + 2*exp(2x) + 2^4*exp(4x)*x^2/2! + 2^9*exp(8x)*x^3/3! +...+ 2^(n^2)*exp(2^n*x)*x^n/n! +...
This is a special case of the more general statement:
Sum_{n>=0} m^n * F(q^n*x)^b * log( F(q^n*x) )^n / n! = Sum_{n>=0} x^n * [y^n] F(y)^(m*q^n + b) where F(x) = exp(x), q=2, m=1, b=1.
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[(2^n+1)^n: n in [0..45]]; // Vincenzo Librandi, Apr 21 2011
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seq((2^n+1)^n, n=0..30); # Robert Israel, Nov 27 2015
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Table[(2^n+1)^n,{n,0,16}] (* Vladimir Joseph Stephan Orlovsky, Feb 14 2011*)
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a(n)=polcoeff(sum(k=0,n,2^(k^2)*exp(2^k*x)*x^k/k!),n)
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{a(n)=polcoeff(sum(k=0, n, 2^(k^2)*x^k/(1-2^k*x +x*O(x^n))^(k+1)), n)} \\ Paul D. Hanna, Sep 15 2009
A244754
a(n) = Sum_{k=0..n} C(n,k) * (1 + 2^k)^(n-k).
Original entry on oeis.org
1, 3, 11, 51, 311, 2583, 30011, 495771, 11740271, 399511023, 19531952051, 1369534859091, 137461591250951, 19708614005005383, 4029559971566918891, 1172950335844577723211, 485515762655939377001951, 285459356061242116657495263, 238215406681004045293498284131
Offset: 0
E.g.f.: A(x) = 1 + 3*x + 11*x^2/2! + 51*x^3/3! + 311*x^4/4! + 2583*x^5/5! +...
where
A(x) = exp(x)*(1 + 2*x + 6*x^2/2! + 26*x^3/3! + 162*x^4/4! + 1442*x^5/5! +...+ A047863(n)*x^n/n! +...).
ILLUSTRATION OF INITIAL TERMS:
a(1) = (1+2^0)^1 + (1+2^1)^0 = 3;
a(2) = (1+2^0)^2 + 2*(1+2^1)^1 + (1+2^2)^0 = 11;
a(3) = (1+2^0)^3 + 3*(1+2^1)^2 + 3*(1+2^2)^1 + (1+2^3)^0 = 51;
a(4) = (1+2^0)^4 + 4*(1+2^1)^3 + 6*(1+2^2)^2 + 4*(1+2^3)^1 + (1+2^4)^0 = 311; ...
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Table[Sum[Binomial[n,k] * (1 + 2^k)^(n-k),{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Jan 25 2015 *)
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{a(n) = sum(k=0,n,binomial(n,k) * (1 + 2^k)^(n-k) )}
for(n=0,25,print1(a(n),", "))
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/* E.g.f.: Sum_{n>=0} exp((1+2^n)*x)*x^n/n!: */
{a(n)=n!*polcoeff(sum(k=0, n, exp((1+2^k)*x +x*O(x^n))*x^k/k!), n)}
for(n=0,25,print1(a(n),", "))
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/* O.g.f. Sum_{n>=0} x^n/(1 - (1+2^n)*x)^(n+1): */
{a(n)=polcoeff(sum(k=0, n, x^k/(1-(1+2^k)*x +x*O(x^n))^(k+1)), n)}
for(n=0,25,print1(a(n),", "))
A244755
a(n) = Sum_{k=0..n} C(n,k) * (1 + 3^k)^(n-k).
Original entry on oeis.org
1, 3, 13, 87, 985, 19563, 697573, 44195007, 4985202865, 987432857043, 344306650353853, 209169206074748967, 222262777197258910345, 409907753371580011362363, 1317924525238880964004945813, 7341603216747343890845790989967, 71176841502529490992224798115792225
Offset: 0
E.g.f.: A(x) = 1 + 3*x + 13*x^2/2! + 87*x^3/3! + 985*x^4/4! + 19563*x^5/5! +...
ILLUSTRATION OF INITIAL TERMS:
a(1) = (1+3^0)^1 + (1+3^1)^0 = 3;
a(2) = (1+3^0)^2 + 2*(1+3^1)^1 + (1+3^2)^0 = 13;
a(3) = (1+3^0)^3 + 3*(1+3^1)^2 + 3*(1+3^2)^1 + (1+3^3)^0 = 87;
a(4) = (1+3^0)^4 + 4*(1+3^1)^3 + 6*(1+3^2)^2 + 4*(1+3^3)^1 + (1+3^4)^0 = 985; ...
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Table[Sum[Binomial[n,k] * (1 + 3^k)^(n-k),{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Jan 25 2015 *)
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{a(n) = sum(k=0,n,binomial(n,k) * (1 + 3^k)^(n-k) )}
for(n=0,25,print1(a(n),", "))
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/* E.g.f. Sum_{n>=0} exp((1+3^n)*x)*x^n/n!" */
{a(n)=n!*polcoeff(sum(k=0, n, exp((1+3^k)*x +x*O(x^n))*x^k/k!), n)}
for(n=0,25,print1(a(n),", "))
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/* O.g.f. Sum_{n>=0} x^n/(1 - (1+3^n)*x)^(n+1): */
{a(n)=polcoeff(sum(k=0, n, x^k/(1-(1+3^k)*x +x*O(x^n))^(k+1)), n)}
for(n=0,25,print1(a(n),", "))
A244756
a(n) = Sum_{k=0..n} C(n,k) * (2 + 3^k)^(n-k).
Original entry on oeis.org
1, 4, 20, 136, 1424, 25504, 831680, 49526656, 5359464704, 1033951896064, 354410768092160, 213011725510260736, 224795751647646224384, 412813583857427719266304, 1323683536183041967893954560, 7361415226356149639592083685376, 71294465534894253722438522191806464
Offset: 0
E.g.f.: A(x) = 1 + 4*x + 20*x^2/2! + 136*x^3/3! + 1424*x^4/4! + 25504*x^5/5! +...
ILLUSTRATION OF INITIAL TERMS:
a(1) = (2+3^0)^1 + (2+3^1)^0 = 4;
a(2) = (2+3^0)^2 + 2*(2+3^1)^1 + (2+3^2)^0 = 20;
a(3) = (2+3^0)^3 + 3*(2+3^1)^2 + 3*(2+3^2)^1 + (2+3^3)^0 = 136;
a(4) = (2+3^0)^4 + 4*(2+3^1)^3 + 6*(2+3^2)^2 + 4*(2+3^3)^1 + (2+3^4)^0 = 1424; ...
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Table[Sum[Binomial[n,k] * (2 + 3^k)^(n-k),{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Jan 25 2015 *)
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{a(n) = sum(k=0,n,binomial(n,k) * (2 + 3^k)^(n-k) )}
for(n=0,25,print1(a(n),", "))
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/* E.g.f. Sum_{n>=0} exp((2+3^n)*x)*x^n/n!" */
{a(n)=n!*polcoeff(sum(k=0, n, exp((2+3^k)*x +x*O(x^n))*x^k/k!), n)}
for(n=0,25,print1(a(n),", "))
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/* O.g.f. Sum_{n>=0} x^n/(1 - (2+3^n)*x)^(n+1): */
{a(n)=polcoeff(sum(k=0, n, x^k/(1-(2+3^k)*x +x*O(x^n))^(k+1)), n)}
for(n=0,25,print1(a(n),", "))
A244760
a(n) = Sum_{k=0..n} C(n,k) * (1 + 3^k)^(n-k) * 2^k.
Original entry on oeis.org
1, 4, 24, 232, 3840, 111904, 5785344, 529662592, 85449338880, 24204383609344, 11986829259362304, 10361640102119729152, 15589910824599107174400, 40815393380277274447519744, 185575767151388880816233447424, 1465910356757779350231777997914112
Offset: 0
E.g.f.: A(x) = 1 + 4*x + 24*x^2/2! + 232*x^3/3! + 3840*x^4/4! + 111904*x^5/5! +...
ILLUSTRATION OF INITIAL TERMS:
a(1) = (1+3^0)^1 + (1+3^1)^0*2 = 4;
a(2) = (1+3^0)^2 + 2*(1+3^1)^1*2 + (1+3^2)^0*2^2 = 24;
a(3) = (1+3^0)^3 + 3*(1+3^1)^2*2 + 3*(1+3^2)^1*2^2 + (1+3^3)^0*2^3 = 232;
a(4) = (1+3^0)^4 + 4*(1+3^1)^3*2 + 6*(1+3^2)^2*2^2 + 4*(1+3^3)^1*2^3 + (1+3^4)^0*2^4 = 3480; ...
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Table[Sum[Binomial[n,k] * (1 + 3^k)^(n-k) * 2^k,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Jan 25 2015 *)
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{a(n) = sum(k=0,n,binomial(n,k) * (1 + 3^k)^(n-k)*2^k )}
for(n=0,25,print1(a(n),", "))
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/* E.g.f. Sum_{n>=0} exp((1+3^n)*x)*(2*x)^n/n! */
{a(n)=n!*polcoeff(sum(k=0, n, exp((1+3^k)*x +x*O(x^n))*(2*x)^k/k!), n)}
for(n=0,25,print1(a(n),", "))
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/* O.g.f. Sum_{n>=0} (2*x)^n/(1 - (1+3^n)*x)^(n+1): */
{a(n)=polcoeff(sum(k=0, n, (2*x)^k/(1-(1+3^k)*x +x*O(x^n))^(k+1)), n)}
for(n=0,25,print1(a(n),", "))
A244004
a(n) = Sum_{k=0..n} binomial(n,k) * (1 + 3^k)^k.
Original entry on oeis.org
1, 5, 109, 22265, 45300601, 865092893645, 151339416167296549, 240067372975786948895105, 3437874756032842865408439504241, 443629315988897178081546025182658298645, 515464811455682924553846499988261249729439172189
Offset: 0
O.g.f.: A(x) = 1 + 5*x + 109*x^2 + 22265*x^3 + 45300601*x^4 +...
where the g.f. may be expressed by the series identity:
A(x) = 1/(1-x) + 4*x/(1-x)^2 + 10^2*x^2/(1-x)^3 + 28^3*x^3/(1-x)^4 + 82^4*x^4/(1-x)^5 + 244^5*x^5/(1-x)^6 + 730^6*x^6/(1-x)^7 +...
A(x) = 1/(1-2*x) + 3*x/(1-4*x)^2 + 3^4*x^2/(1-10*x)^3 + 3^9*x^3/(1-28*x)^4 + 3^16*x^4/(1-82*x)^5 + 3^25*x^5/(1-244*x)^6 + 3^36*x^6/(1-730*x)^7 +...
Illustration of initial terms:
a(0) = 1;
a(1) = 1 + (1+3);
a(2) = 1 + 2*(1+3) + (1+3^2)^2;
a(3) = 1 + 3*(1+3) + 3*(1+3^2)^2 + (1+3^3)^3;
a(4) = 1 + 4*(1+3) + 6*(1+3^2)^2 + 4*(1+3^3)^3 + (1+3^4)^4;
a(5) = 1 + 5*(1+3) + 10*(1+3^2)^2 + 10*(1+3^3)^3 + 5*(1+3^4)^4 + (1+3^5)^5; ...
Also, by a binomial identity we have
a(0) = 1;
a(1) = 2 + 3;
a(2) = 2^2 + 2*(1+3)*3 + 3^4;
a(3) = 2^3 + 3*(1+3)^2*3 + 3*(1+3^2)*3^4 + 3^9;
a(4) = 2^4 + 4*(1+3)^3*3 + 6*(1+3^2)^2*3^4 + 4*(1+3^3)*3^9 + 3^16;
a(5) = 2^5 + 5*(1+3)^4*3 + 10*(1+3^2)^3*3^4 + 10*(1+3^3)^2*3^9 + 5*(1+3^4)*3^16 + 3^25; ...
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Table[Sum[Binomial[n,k]*(1+3^k)^k,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Jun 18 2014 *)
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{a(n)=sum(k=0, n, binomial(n, k)*(1+3^k)^k)}
for(n=0, 20, print1(a(n), ", "))
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{a(n)=sum(k=0, n, binomial(n, k)*(1+3^k)^(n-k)*3^(k^2))}
for(n=0, 20, print1(a(n), ", "))
Showing 1-6 of 6 results.
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