A138247 -id:A138247 - cp's OEIS frontend

A133049 Squares of Mersenne primes A000668(n).

9, 49, 961, 16129, 67092481, 17179607041, 274876858369, 4611686014132420609, 5316911983139663487003542222693990401, 383123885216472214589586755549637256619304505646776321

Offset: 1

Omar E. Pol, Oct 30 2007, Apr 23 2008

nonn

Sum of last A000043(n) divisors of the n-th even perfect number. In other words; sum of divisors that are not powers of 2 of the n-th even perfect number, or sum of divisors that are multiples of the n-th Mersenne prime A000668(n) of the n-th even perfect number. See A139247 for more information.

See the structure of the divisors of perfect numbers in A135652, A135653, A135654 and A135655.

			a(3)=961 because the 3rd Mersenne prime is 31 and 31^2=961.

G. C. Greubel, Table of n, a(n) for n = 1..15
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.

Cf. A000290, A001248. Mersenne primes: A000668.

Cf. A000043, A000396, A135652, A135653, A135654, A135655, A138247.

Select[2^Range[1000] - 1, PrimeQ]^2 (* G. C. Greubel, Oct 03 2017 *)

forprime(p=2, 1000, if(ispseudoprime(2^p-1), print1((2^p-1)^2", "))) \\ G. C. Greubel, Oct 03 2017

a(n) = A000668(n)^2

More terms from Olaf Voß, Feb 13 2008

A094617 Triangular array T of numbers generated by these rules: 2 is in T; and if x is in T, then 2x-1 and 3x-2 are in T.

Original entry on oeis.org

2, 3, 4, 5, 7, 10, 9, 13, 19, 28, 17, 25, 37, 55, 82, 33, 49, 73, 109, 163, 244, 65, 97, 145, 217, 325, 487, 730, 129, 193, 289, 433, 649, 973, 1459, 2188, 257, 385, 577, 865, 1297, 1945, 2917, 4375, 6562, 513, 769, 1153, 1729, 2593, 3889, 5833, 8749, 13123, 19684

Offset: 1

Clark Kimberling, May 14 2004

To obtain row n from row n-1, apply 2x-1 to each x in row n-1 and then put 1+3^n at the end. Or, instead, apply 3x-2 to each x in row n-1 and then put 1+2^n at the beginning.
From Lamine Ngom, Feb 10 2021: (Start)
Triangle read by diagonals provides all the sequences of the form 1+2^(k-1)*3^n, where k is the k-th diagonal.
For instance, the terms of the first diagonal form the sequence 2, 4, 10, 28, ..., i.e., 1+3^n (A034472).
The 2nd diagonal leads to the sequence 3, 7, 19, 55, ..., i.e., 1+2*3^n (A052919).
The 3rd diagonal is the sequence 5, 13, 37, 109, ..., i.e., 1+4*3^n (A199108).
And for k = 4, we obtain the sequence 9, 25, 73, 217, ..., i.e., 1+8*3^n (A199111). (End)

			Rows of this triangle begin:
    2;
    3,   4;
    5,   7,   10;
    9,  13,   19,   28;
   17,  25,   37,   55,   82;
   33,  49,   73,  109,  163,  244;
   65,  97,  145,  217,  325,  487,  730;
  129, 193,  289,  433,  649,  973, 1459, 2188;
  257, 385,  577,  865, 1297, 1945, 2917, 4375,  6562;
  513, 769, 1153, 1729, 2593, 3889, 5833, 8749, 13123, 19684;
  ...

Ivan Neretin, Table of n, a(n) for n = 1..5151

Cf. A094616, A094618, A138247.

Cf. A034472, A052919, A199108, A199111.

FoldList[Append[2 #1 - 1, 1 + 3^#2] &, {2}, Range[9]] // Flatten (* Ivan Neretin, Mar 30 2016 *)

When offset is zero, then the first term is T(0,0) = 2, and
T(n,0) = 1 + 2^n = A000051(n),
T(n,n) = 1 + 3^n = A048473(n),
T(2n,n) = 1 + 6^n = A062394(n).
Row sums = A094618.
a(n) = A036561(n-1) + 1. - Filip Zaludek, Nov 19 2016

A196460 E.g.f.: A(x) = Sum_{n>=0} (1+2^n)^n * exp((1+2^n)x) x^n/n!.

Original entry on oeis.org

1, 5, 47, 1193, 113855, 46857665, 83540629607, 629692415941433, 19653639560140008575, 2505063418700072099312705, 1292764583816731772891346438887, 2687238342732260436646020885678131993, 22431974111110989403331425804893720873764255

Offset: 0

Paul D. Hanna, Oct 02 2011

nonn

GENERAL BINOMIAL IDENTITY.
When p=1, q=2, this sequence illustrates the following identity.
Given e.g.f.: Sum_{n>=0} (p^n+q^n)^n*exp((p^n+q^n)*x)*x^n/n! = Sum_{n>=0} a(n)*x^n/n!,
then a(n) = Sum_{k=0..n} C(n,k)*(p^k + q^k)^n = Sum_{k=0..n} C(n,k)*(1 + p^(n-k)*q^k)^n;
which is a special case of the more general binomial identity:
Sum_{k=0..n} C(n,k)*(s*p^k + t*q^k)^(n-k) * (u*p^k + v*q^k)^k = Sum_{k=0..n} C(n,k)*(s + u*p^(n-k)*q^k)^(n-k) * (t + v*p^(n-k)*q^k)^k.

			E.g.f.: A(x) = 1 + 5*x + 47*x^2/2! + 1193*x^3/3! + 113855*x^4/4! +...
where
A(x) = exp((1+1)*x) + (1+2)*exp((1+2)*x)*x + (1+2^2)^2*exp((1+2^2)*x)*x^2/2! + (1+2^3)^3*exp((1+2^3)*x)*x^3/3! +...
or, equivalently,
A(x) = exp(2*x) + 3*exp(3*x)*x + 5^2*exp(5*x)*x^2/2! + 9^3*exp(9*x)*x^3/3! + 17^4*exp(17*x)*x^4/4! + 33^5*exp(33*x)*x^5/5! +...
Illustrate the formula for the terms:
a(1) = (1+1) + (1+2) = 5 ;
a(2) = (1+1)^2 + 2*(1+2)^2 + (1+2^2)^2 = 2^2 + 2*3^2 + 5^2 = 47 ;
a(3) = (1+1)^3 + 3*(1+2)^3 + 3*(1+2^2)^3 + (1+2^3)^3 = 2^3 + 3*3^3 + 3*5^3 + 9^3 = 1193 ;
a(4) = (1+1)^4 + 4*(1+2)^4 + 6*(1+2^2)^4 + 4*(1+2^3)^4 + (1+2^4)^4 = 2^4 + 4*3^4 + 6*5^4 + 4*9^4 + 17^4 = 113855.

Cf. A138247, A196457.

Table[Sum[Binomial[n,k]*(1+2^k)^n, {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Jun 25 2013 *)

{a(n)=local(p=1, q=2);n!*polcoeff(sum(m=0,n,(p^m+q^m)^m*exp((p^m+q^m+x*O(x^n))*x)*x^m/m!),n)}

{a(n)=local(p=1, q=2, s=1, t=1, u=1, v=1);
sum(k=0, n, binomial(n, k)*(s*p^k + t*q^k)^(n-k)*(u*p^k + v*q^k)^k)}

/* right side of the general binomial identity: */
{a(n)=local(p=1, q=2, s=1, t=1, u=1, v=1);
sum(k=0, n, binomial(n, k)*(s + u*p^(n-k)*q^k)^(n-k) * (t + v*p^(n-k)*q^k)^k)}

GENERATING FUNCTIONS.
E.g.f.: Sum_{n>=0} (1 + 2^n)^n * exp( (1 + 2^n)*x ) * x^n / n!.
O.g.f.: Sum_{n>=0} (1 + 2^n)^n * x^n / (1 - (1 + 2^n)*x)^(n+1). - Paul D. Hanna, Jul 13 2019
FORMULAS FOR TERMS.
a(n) = Sum_{k=0..n} binomial(n,k) * (1 + 2^k)^n.
a(n) ~ 2^(n^2). - Vaclav Kotesovec, Jun 25 2013

A196457 E.g.f.: A(x) = Sum_{n>=0} exp((2^n + (-1)^n)x) (2^n + (-1)^n)^n * x^n/n!.

Original entry on oeis.org

1, 3, 31, 729, 96895, 35927793, 81108563671, 567783612614529, 19581520178825073535, 2420011073132910603900513, 1292280969200128366004695992151, 2658679109878459106807828064662797809, 22431208469091982323298987880694649428158815, 748294346623782293365235855701111498805828889778353

Offset: 0

Paul D. Hanna, Oct 03 2011

nonn

GENERAL BINOMIAL IDENTITY.
When p=-1, q=2, this sequence illustrates the following identity.
Given e.g.f.: Sum_{n>=0} (p^n+q^n)^n*exp((p^n+q^n)*x)*x^n/n! = Sum_{n>=0} a(n)*x^n/n!,
then a(n) = Sum_{k=0..n} C(n,k)*(p^k + q^k)^n =  Sum_{k=0..n} C(n,k)*(1 + p^(n-k)*q^k)^n;
which is a special case of the more general binomial identity:
Sum_{k=0..n} C(n,k)*(s*p^k + t*q^k)^(n-k) * (u*p^k + v*q^k)^k = Sum_{k=0..n} C(n,k)*(s + u*p^(n-k)*q^k)^(n-k) * (t + v*p^(n-k)*q^k)^k.

			E.g.f.: A(x) = 1 + 3*x + 31*x^2/2! + 729*x^3/3! + 96895*x^4/4! +...
where
A(x) = exp((1+1)*x) + (2-1)*exp((2-1)*x)*x + (2^2+1)^2*exp((2^2+1)*x)*x^2/2! + (2^3-1)^3*exp((2^3-1)*x)*x^3/3! +...
or, equivalently,
A(x) = exp(2*x) + 1*exp(1*x)*x + 5^2*exp(5*x)*x^2/2! + 7^3*exp(7*x)*x^3/3! + 17^4*exp(17*x)*x^4/4! + 31^5*exp(31*x)*x^5/5! +...
Illustrate the formula for the terms:
a(1) = (1+1) + (2-1) = 3 ;
a(2) = (1+1)^2 + 2*(2-1)^2 + (2^2+1)^2 = 2^2 + 2*1^2 + 5^2 = 31 ;
a(3) = (1+1)^3 + 3*(2-1)^3 + 3*(2^2+1)^3 + (2^3-1)^3 = 2^3 + 3*1^3 + 3*5^3 + 7^3 = 729 ;
a(4) = (1+1)^4 + 4*(2-1)^4 + 6*(2^2+1)^4 + 4*(2^3-1)^4 + (2^4+1)^4 = 2^4 + 4*1^4 + 6*5^4 + 4*7^4 + 17^4 = 96895.

Cf. A138247, A196458, A196459, A196460.

{a(n)=n!*polcoeff(sum(m=0,n,exp((2^m+(-1)^m+x*O(x^n))*x)*(2^m+(-1)^m)^m*x^m/m!),n)}

{a(n)=sum(k=0,n,binomial(n,k)*(2^k + (-1)^k)^n)}

{a(n)=local(p=-1, q=2);n!*polcoeff(sum(m=0,n,(p^m+q^m)^m*exp((p^m+q^m+x*O(x^n))*x)*x^m/m!),n)}

{a(n)=local(p=-1, q=2, s=1, t=1, u=1, v=1);
sum(k=0, n, binomial(n, k)*(s*p^k + t*q^k)^(n-k)*(u*p^k + v*q^k)^k)}

/* right side of the general binomial identity: */
{a(n)=local(p=-1, q=2, s=1, t=1, u=1, v=1);
sum(k=0, n, binomial(n, k)*(s + u*p^(n-k)*q^k)^(n-k) * (t + v*p^(n-k)*q^k)^k)}

GENERATING FUNCTIONS.
E.g.f.: Sum_{n>=0} (2^n + (-1)^n)^n * exp( (2^n + (-1)^n)*x ) * x^n/n!.
O.g.f.: Sum_{n>=0} (2^n + (-1)^n)^n * x^n / (1 - (2^n + (-1)^n)*x)^(n+1). - Paul D. Hanna, Jul 13 2019
FORMULAS FOR TERMS.
a(n) = Sum_{k=0..n} C(n,k)*(2^k + (-1)^k)^n.

A196459 E.g.f.: A(x) = Sum_{n>=0} (2^n + 3^n)^n * exp((52^n + 23^n)x) x^n/n!.

Original entry on oeis.org

1, 12, 378, 66324, 106198818, 1683766925772, 254853525616593498, 359442643592845468030044, 4678184388343291088594901552738, 559325487076698590861626663741490993292, 612293179823760898820162678475549198446848819338

Offset: 0

Paul D. Hanna, Oct 04 2011

nonn

This sequence illustrates the following general binomial identity.
Given e.g.f.: Sum_{n>=0} exp((s*p^n + t*q^n)*x) * (u*p^n + v*q^n)^n * x^n/n! = Sum_{n>=0} a(n)*x^n/n!,
then a(n) = Sum_{k=0..n} C(n,k)*(s*p^k + t*q^k)^(n-k) * (u*p^k + v*q^k)^k = Sum_{k=0..n} C(n,k)*(s + u*p^(n-k)*q^k)^(n-k) * (t + v*p^(n-k)*q^k)^k.

			E.g.f.: A(x) = 1 + 12*x + 378*x^2/2! + 66324*x^3/3! + 106198818*x^4/4! +...
where
A(x) = exp((5+2)*x) + (2+3)*exp((5*2+2*3)*x)*x + (2^2+3^2)^2*exp((5*2^2+2*3^2)*x)*x^2/2! + (2^3+3^3)^3*exp((5*2^3+2*3^3)*x)*x^3/3! +...
or, equivalently,
A(x) = exp(7*x) + 5*exp(16*x)*x + 13^2*exp(38*x)*x^2/2! + 35^3*exp(94*x)*x^3/3! + 97^4*exp(242*x)*x^4/4! + 275^5*exp(646*x)*x^5/5! +...
Illustrate formula (1):
a(1) = 7 + 5 = 12 ;
a(2) = 7^2 + 2*5*16 + 13^2 = 378 ;
a(3) = 7^3 + 3*5*16^2 + 3*13^2*38 + 35^3 = 66324 ;
a(4) = 7^4 + 4*5*16^3 + 6*13^2*38^2 + 4*35^3*94 + 97^4 = 106198818 ;
a(5) = 7^5 + 5*5*16^4 + 10*13^2*38^3 + 10*35^3*94^2 + 5*97^4*242 + 275^5 = 1683766925772; ...
Illustrate formula (2):
a(1) = 7 + 5 = 12 ;
a(2) = 9^2 + 2*11*8 + 14*11^2 = 378 ;
a(3) = 13^3 + 3*17^2*14 + 3*23*20^2 + 29^3 = 66324 ;
a(4) = 21^4 + 4*29^3*26 + 6*41^2*38^2 + 4*59*56^3 + 83^4 = 106198818 ;
a(5) = 37^5 + 5*53^4*50 + 10*77^3*74^2 + 10*113^2*110^3 + 5*167*164^4 + 1*245^5 = 1683766925772; ...

Cf. A196460, A138247.

{a(n)=n!*polcoeff(sum(m=0,n,exp((5*2^m+2*3^m+x*O(x^n))*x)*(2^m+3^m)^m*x^m/m!),n)}

{a(n)=sum(k=0,n,binomial(n,k)*(5*2^k + 2*3^k)^(n-k)*(2^k + 3^k)^k)}

{a(n)=sum(k=0,n,binomial(n,k)*(5 + 2^(n-k)*3^k)^(n-k)*(2 + 2^(n-k)*3^k)^k)}

{a(n)=local(p=2, q=3,s=5,t=2,u=1,v=1);n!*polcoeff(sum(m=0,n,exp((s*p^m+t*q^m+x*O(x^n))*x)*(u*p^m+v*q^m)^m*x^m/m!),n)}

{a(n)=local(p=2, q=3, s=5, t=2, u=1, v=1);
sum(k=0, n, binomial(n, k)*(s*p^k + t*q^k)^(n-k)*(u*p^k + v*q^k)^k)}

/* right side of the general binomial identity: */
{a(n)=local(p=2, q=3, s=5, t=2, u=1, v=1);
sum(k=0, n, binomial(n, k)*(s + u*p^(n-k)*q^k)^(n-k) * (t + v*p^(n-k)*q^k)^k)}

(1) a(n) = Sum_{k=0..n} C(n,k)*(5*2^k + 2*3^k)^(n-k)*(2^k + 3^k)^k.

(2) a(n) = Sum_{k=0..n} C(n,k)*(5 + 2^(n-k)*3^k)^(n-k)*(2 + 2^(n-k)*3^k)^k.

A196458 E.g.f.: A(x) = Sum_{n>=0} (3^n + (-1)^n)^n * exp((3^n + (-1)^n)x) x^n/n!.

Original entry on oeis.org

1, 4, 112, 20608, 47100160, 848654393344, 152543949079048192, 239308785705492230176768, 3442046584639832610980531077120, 443426848780270385458655031044167696384, 515552048984399455145742768443316759297510080512

Offset: 0

Paul D. Hanna, Nov 20 2011

nonn

 When p=-1, q=3, this sequence illustrates the following identity.
Given e.g.f.: Sum_{n>=0} (p^n+q^n)^n*exp((p^n+q^n)*x)*x^n/n! = Sum_{n>=0} a(n)*x^n/n!,
then a(n) = Sum_{k=0..n} C(n,k)*(p^k + q^k)^n =  Sum_{k=0..n} C(n,k)*(1 + p^(n-k)*q^k)^n;
which is a special case of the more general binomial identity:
Sum_{k=0..n} C(n,k)*(s*p^k + t*q^k)^(n-k) * (u*p^k + v*q^k)^k = Sum_{k=0..n} C(n,k)*(s + u*p^(n-k)*q^k)^(n-k) * (t + v*p^(n-k)*q^k)^k.

			 E.g.f.: A(x) = 1 + 4*x + 112*x^2/2! + 20608*x^3/3! + 47100160*x^4/4! +...
where
_ A(x) = exp((1+1)*x) + (3-1)*exp((3-1)*x)*x + (3^2+1)^2*exp((3^2+1)*x)*x^2/2! + (3^3-1)^3*exp((3^3-1)*x)*x^3/3! +...
or, equivalently,
_ A(x) = exp(2*x) + 2*exp(2*x)*x + 10^2*exp(10*x)*x^2/2! + 26^3*exp(26*x)*x^3/3! + 82^4*exp(82*x)*x^4/4! + 242^5*exp(242*x)*x^5/5! +...
Illustrate the formula for the terms:
a(1) = (1+1) + (3-1) = 4 ;
a(2) = (1+1)^2 + 2*(3-1)^2 + (3^2+1)^2 = 2^2 + 2*2^2 + 10^2 = 112 ;
a(3) = (1+1)^3 + 3*(3-1)^3 + 3*(3^2+1)^3 + (3^3-1)^3 = 2^3 + 3*2^3 + 3*10^3 + 26^3 = 20608 ;
a(4) = (1+1)^4 + 4*(3-1)^4 + 6*(3^2+1)^4 + 4*(3^3-1)^4 + (3^4+1)^4 = 2^4 + 4*2^4 + 6*10^4 + 4*26^4 + 82^4 = 47100160.

Cf. A196457, A196459, A196460, A138247.

{a(n)=sum(k=0,n,binomial(n,k)*(3^k + (-1)^k)^n)}

{a(n)=sum(k=0,n,binomial(n,k)*(1 + (-1)^(n-k)*3^k)^n)}

{a(n)=local(p=-1, q=3);n!*polcoeff(sum(m=0,n,(p^m+q^m)^m*exp((p^m+q^m+x*O(x^n))*x)*x^m/m!),n)}

{a(n)=local(p=-1, q=3, s=1, t=1, u=1, v=1);
sum(k=0, n, binomial(n, k)*(s*p^k + t*q^k)^(n-k)*(u*p^k + v*q^k)^k)}

/* right side of the general binomial identity: */
{a(n)=local(p=-1, q=3, s=1, t=1, u=1, v=1);
sum(k=0, n, binomial(n, k)*(s + u*p^(n-k)*q^k)^(n-k) * (t + v*p^(n-k)*q^k)^k)}

a(n) = Sum_{k=0..n} C(n,k)*(3^k + (-1)^k)^n.

a(n) = Sum_{k=0..n} C(n,k)*(1 + (-1)^(n-k)*3^k)^n.

A326599 G.f.: Sum_{n>=0} x^n * (1 + x^n)^n / (1 - x*(1 + x^n))^(n+1).

Original entry on oeis.org

1, 3, 8, 19, 44, 97, 207, 432, 884, 1777, 3529, 6942, 13547, 26281, 50791, 97942, 188677, 363489, 700953, 1354060, 2621602, 5088832, 9905382, 19335477, 37848971, 74287855, 146173721, 288280956, 569715036, 1127957956, 2236777539, 4441749653, 8830819362, 17574636239, 35005944165, 69776276002, 139165947494

Offset: 0

Paul D. Hanna, Jul 13 2019

nonn

			G.f.: A(x) = 1 + 3*x + 8*x^2 + 19*x^3 + 44*x^4 + 97*x^5 + 207*x^6 + 432*x^7 + 884*x^8 + 1777*x^9 + 3529*x^10 + 6942*x^11 + 13547*x^12 + 26281*x^13 + ...
such that
A(x) = 1/(1-2*x) + x*(1+x)/(1 - x*(1+x))^2 + x^2*(1+x^2)^2/(1 - x*(1+x^2))^3 + x^3*(1+x^3)^3/(1 - x*(1+x^3))^4 + x^4*(1+x^4)^4/(1 - x*(1+x^4))^5  + x^5*(1+x^5)^5/(1 - x*(1+x^5))^6 + x^6*(1+x^6)^6/(1 - x*(1+x^6))^7 + ...

Cf. A138247.

{a(n) = my(A = sum(m=0,n, x^m * (1 + x^m +x*O(x^n))^m /(1 - x*(1+x^m)  +x*O(x^n))^(m+1) )); polcoeff(A,n)}
for(n=0,50,print1(a(n),", "))

cp's OEIS Frontend

A133049 Squares of Mersenne primes A000668(n).

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A094617 Triangular array T of numbers generated by these rules: 2 is in T; and if x is in T, then 2x-1 and 3x-2 are in T.

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A196460 E.g.f.: A(x) = Sum_{n>=0} (1+2^n)^n * exp((1+2^n)*x) * x^n/n!.

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A196457 E.g.f.: A(x) = Sum_{n>=0} exp((2^n + (-1)^n)*x) * (2^n + (-1)^n)^n * x^n/n!.

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A196459 E.g.f.: A(x) = Sum_{n>=0} (2^n + 3^n)^n * exp((5*2^n + 2*3^n)*x) * x^n/n!.

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A196458 E.g.f.: A(x) = Sum_{n>=0} (3^n + (-1)^n)^n * exp((3^n + (-1)^n)*x) * x^n/n!.

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PARI

PARI

A196460 E.g.f.: A(x) = Sum_{n>=0} (1+2^n)^n * exp((1+2^n)x) x^n/n!.

A196457 E.g.f.: A(x) = Sum_{n>=0} exp((2^n + (-1)^n)x) (2^n + (-1)^n)^n * x^n/n!.

A196459 E.g.f.: A(x) = Sum_{n>=0} (2^n + 3^n)^n * exp((52^n + 23^n)x) x^n/n!.

A196458 E.g.f.: A(x) = Sum_{n>=0} (3^n + (-1)^n)^n * exp((3^n + (-1)^n)x) x^n/n!.