cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 31 results. Next

A026898 a(n) = Sum_{k=0..n} (n-k+1)^k.

Original entry on oeis.org

1, 2, 4, 9, 23, 66, 210, 733, 2781, 11378, 49864, 232769, 1151915, 6018786, 33087206, 190780213, 1150653921, 7241710930, 47454745804, 323154696185, 2282779990495, 16700904488706, 126356632390298, 987303454928973, 7957133905608837, 66071772829247410
Offset: 0

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Author

Keywords

Comments

Row sums of A004248, A009998, A009999.
First differences are in A047970.
First differences of A103439.
Antidiagonal sums of array A003992.
a(n-1), for n>=1, is the number of length-n restricted growth strings (RGS) [s(0),s(1),...,s(n-1)] where s(0)=0 and s(k)<=1+max(prefix) for k>=1, that are simultaneously projections as maps f: [n] -> [n] where f(x)<=x and f(f(x))=f(x); see example and the two comments (Arndt, Apr 30 2011 Jan 04 2013) in A000110. - Joerg Arndt, Mar 07 2015
Number of finite sequences s of length n+1 whose discriminator sequence is s itself. Here the discriminator sequence of s is the one where the n-th term (n>=1) is the least positive integer k such that the first n terms are pairwise incongruent, modulo k. - Jeffrey Shallit, May 17 2016
From Gus Wiseman, Jan 08 2019: (Start)
Also the number of set partitions of {1,...,n+1} whose minima form an initial interval of positive integers. For example, the a(3) = 9 set partitions are:
{{1},{2},{3},{4}}
{{1},{2},{3,4}}
{{1},{2,4},{3}}
{{1,4},{2},{3}}
{{1},{2,3,4}}
{{1,3},{2,4}}
{{1,4},{2,3}}
{{1,3,4},{2}}
{{1,2,3,4}}
Missing from this list are:
{{1},{2,3},{4}}
{{1,2},{3},{4}}
{{1,3},{2},{4}}
{{1,2},{3,4}}
{{1,2,3},{4}}
{{1,2,4},{3}}
(End)
a(n) is the number of m-tuples of nonnegative integers less than or equal to n-m (including the "0-tuple"). - Mathew Englander, Apr 11 2021

Examples

			G.f.: A(x) = 1 + 2*x + 4*x^2 + 9*x^3 + 23*x^4 + 66*x^5 + 210*x^6 + ...
where we have the identity:
A(x) = 1/(1-x) + x/(1-2*x) + x^2/(1-3*x) + x^3/(1-4*x) + x^4/(1-5*x) + ...
is equal to
A(x) = 1/(1-x) + x/((1-x)^2*(1+x)) + 2!*x^2/((1-x)^3*(1+x)*(1+2*x)) + 3!*x^3/((1-x)^4*(1+x)*(1+2*x)*(1+3*x)) + 4!*x^4/((1-x)^5*(1+x)*(1+2*x)*(1+3*x)*(1+4*x)) + ...
From _Joerg Arndt_, Mar 07 2015: (Start)
The a(5-1) = 23 RGS described in the comment are (dots denote zeros):
01:  [ . . . . . ]
02:  [ . 1 . . . ]
03:  [ . 1 . . 1 ]
04:  [ . 1 . 1 . ]
05:  [ . 1 . 1 1 ]
06:  [ . 1 1 . . ]
07:  [ . 1 1 . 1 ]
08:  [ . 1 1 1 . ]
09:  [ . 1 1 1 1 ]
10:  [ . 1 2 . . ]
11:  [ . 1 2 . 1 ]
12:  [ . 1 2 . 2 ]
13:  [ . 1 2 1 . ]
14:  [ . 1 2 1 1 ]
15:  [ . 1 2 1 2 ]
16:  [ . 1 2 2 . ]
17:  [ . 1 2 2 1 ]
18:  [ . 1 2 2 2 ]
19:  [ . 1 2 3 . ]
20:  [ . 1 2 3 1 ]
21:  [ . 1 2 3 2 ]
22:  [ . 1 2 3 3 ]
23:  [ . 1 2 3 4 ]
(End)
		

Crossrefs

Programs

  • Haskell
    a026898 n = sum $ zipWith (^) [n + 1, n .. 1] [0 ..]
    -- Reinhard Zumkeller, Sep 14 2014
    
  • Magma
    [(&+[(n-k+1)^k: k in [0..n]]): n in [0..50]]; // Stefano Spezia, Jan 09 2019
    
  • Maple
    a:= n-> add((n+1-j)^j, j=0..n): seq(a(n), n=0..23); # Zerinvary Lajos, Apr 18 2009
  • Mathematica
    Table[Sum[(n-k+1)^k, {k,0,n}], {n, 0, 25}] (* Michael De Vlieger, Apr 01 2015 *)
  • PARI
    {a(n)=polcoeff(sum(m=0,n,x^m/(1-(m+1)*x+x*O(x^n))),n)} /* Paul D. Hanna, Sep 13 2011 */
    
  • PARI
    {INTEGRATE(n,F)=local(G=F);for(i=1,n,G=intformal(G));G}
    {a(n)=local(A=1+x);A=sum(k=0,n,INTEGRATE(k,exp((k+1)*x+x*O(x^n))));n!*polcoeff(A,n)} \\ Paul D. Hanna, Dec 28 2013
    for(n=0,30,print1(a(n),", "))
    
  • PARI
    {a(n)=polcoeff( sum(m=0, n, m!*x^m/(1-x +x*O(x^n))^(m+1)/prod(k=1, m, 1+k*x +x*O(x^n))), n)}  /* From o.g.f. (Paul D. Hanna, Jul 20 2014) */
    for(n=0, 25, print1(a(n), ", "))
    
  • Sage
    [sum((n-j+1)^j for j in (0..n)) for n in (0..30)] # G. C. Greubel, Jun 15 2021

Formula

a(n) = A003101(n) + 1.
G.f.: Sum_{n>=0} x^n/(1 - (n+1)*x). - Paul D. Hanna, Sep 13 2011
G.f.: G(0) where G(k) = 1 + x*(2*k*x-1)/((2*k*x+x-1) - x*(2*k*x+x-1)^2/(x*(2*k*x+x-1) + (2*k*x+2*x-1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 26 2013
E.g.f.: Sum_{n>=0} Integral^n exp((n+1)*x) dx^n, where Integral^n F(x) dx^n is the n-th integration of F(x) with no constant of integration. - Paul D. Hanna, Dec 28 2013
O.g.f.: Sum_{n>=0} n! * x^n/(1-x)^(n+1) / Product_{k=1..n} (1 + k*x). - Paul D. Hanna, Jul 20 2014
a(n) = A101494(n+1,0). - Vladimir Kruchinin, Apr 01 2015
a(n-1) = Sum_{k = 1..n} k^(n-k). - Gus Wiseman, Jan 08 2019
log(a(n)) ~ (1 - 1/LambertW(exp(1)*n)) * n * log(1 + n/LambertW(exp(1)*n)). - Vaclav Kotesovec, Jun 15 2021
a(n) ~ sqrt(2*Pi/(n+1 + (n+1)/w(n))) * ((n+1)/w(n))^(n+2 - (n+1)/w(n)), where w(n) = LambertW(exp(1)*(n+1)). - Vaclav Kotesovec, Jun 25 2021, after user "leonbloy", see Mathematics Stack Exchange link.

Extensions

a(23)-a(25) from Paul D. Hanna, Dec 28 2013

A047970 Antidiagonal sums of nexus numbers (A047969).

Original entry on oeis.org

1, 2, 5, 14, 43, 144, 523, 2048, 8597, 38486, 182905, 919146, 4866871, 27068420, 157693007, 959873708, 6091057009, 40213034874, 275699950381, 1959625294310, 14418124498211, 109655727901592, 860946822538675, 6969830450679864, 58114638923638573
Offset: 0

Views

Author

Alford Arnold, Dec 11 1999

Keywords

Comments

From Lara Pudwell, Oct 23 2008: (Start)
A permutation p avoids a pattern q if it has no subsequence that is order-isomorphic to q. For example, p avoids the pattern 132 if it has no subsequence abc with a < c < b.
Barred pattern avoidance considers permutations that avoid a pattern except in a special case. Given a barred pattern q, we may form two patterns, q1 = the sequence of unbarred letters of q and q2 = the sequence of all letters of q.
A permutation p avoids barred pattern q if every instance of q1 in p is embedded in a copy of q2 in p. In other words, p avoids q1, except in the special case that a copy of q1 is a subsequence of a copy of q2.
For example, if q=5{bar 1}32{bar 4}, then q1=532 and q2 = 51324. p avoids q if every for decreasing subsequence acd of length 3 in p, one can find letters b and e so that the subsequence abcde of p has b < d < c < e < a. (End)
Number of ordered factorizations over the Gaussian polynomials.
Apparently, also the number of permutations in S_n avoiding {bar 3}{bar 1}542 (i.e., every occurrence of 542 is contained in an occurrence of a 31542). - Lara Pudwell, Apr 25 2008
With offset 1, apparently the number of sequences {b(m)} of length n of positive integers with b(1) = 1 and, for all m > 1, b(m) <= max{b(m-1) + 1, max{b(i) | 1 <= i <= m - 1}}. This sequence begins 1, 2, 5, 14, 43, 144, 523, 2048, 8597, 38486. The term 144 counts the length 6 sequence 1, 2, 3, 1, 1, 3, for instance. Contrast with the families of sequences discussed in Franklin T. Adams-Watters's comment in A005425. - Rick L. Shepherd, Jan 01 2015
a(n-1) for n >= 1 is the number of length-n restricted growth strings (RGS) [s(0), s(1), ..., s(n-1)] with s(0)=0 and s(k) <= the number of fixed points in the prefix, see example. - Joerg Arndt, Mar 08 2015
Number of sequences (e(1), ..., e(n+1)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) != e(j) = e(k). [Martinez and Savage, 2.15] - Eric M. Schmidt, Jul 17 2017
a(n) counts all positive-integer m-tuples whose maximum is n-m+2. - Mathew Englander, Feb 28 2021
a(n) counts the cyclic permutations of [n+2] that avoid the vincular pattern 12-3-4, i.e., the pattern 1234 where the 1 and 2 are required to be adjacent. - Rupert Li, Jul 27 2021

Examples

			a(3) = 1 + 5 + 7 + 1 = 14.
From _Paul D. Hanna_, Jul 22 2014:  (Start)
G.f. A(x) = 1 + 2*x + 5*x^2 + 14*x^3 + 43*x^4 + 144*x^5 + 523*x^6 + 2048*x^7 + ...
where we have the series identity:
A(x) = (1-x)*( 1/(1-2*x) + x/(1-3*x) + x^2/(1-4*x) + x^3/(1-5*x) + x^4/(1-6*x) + x^5/(1-7*x) + x^6/(1-8*x) + ...)
is equal to
A(x) = 1/(1-x) + x/((1-x)*(1-2*x)) + x^2/((1-2*x)*(1-3*x)) + x^3/((1-3*x)*(1-4*x)) + x^4/((1-4*x)*(1-5*x)) + x^5/((1-5*x)*(1-6*x)) + x^6/((1-6*x)*(1-7*x)) + ...
and also equals
A(x) = 1/((1-x)*(1+x)) + 2!*x/((1-x)^2*(1+x)*(1+2*x)) + 3!*x^2/((1-x)^3*(1+x)*(1+2*x)*(1+3*x)) + 4!*x^3/((1-x)^4*(1+x)*(1+2*x)*(1+3*x)*(1+4*x)) + ...
(End)
From _Joerg Arndt_, Mar 08 2015: (Start)
There are a(4-1)=14 length-4 RGS as in the comment (dots denote zeros):
01:  [ . . . . ]
02:  [ . . . 1 ]
03:  [ . . 1 . ]
04:  [ . . 1 1 ]
05:  [ . 1 . . ]
06:  [ . 1 . 1 ]
07:  [ . 1 . 2 ]
08:  [ . 1 1 . ]
09:  [ . 1 1 1 ]
10:  [ . 1 1 2 ]
11:  [ . 1 2 . ]
12:  [ . 1 2 1 ]
13:  [ . 1 2 2 ]
14:  [ . 1 2 3 ]
(End)
		

Crossrefs

Antidiagonal sums of A085388 (beginning with the second antidiagonal) and A047969.
Partial sums are in A026898, A003101. First differences A112532.

Programs

  • Maple
    T := proc(n, k) option remember; local j;
        if k=n then 1
      elif k>n then 0
      else (k+1)*T(n-1, k) + add(T(n-1, j), j=k..n)
        fi end:
    A047970 := n -> T(n,0);
    seq(A047970(n), n=0..24); # Peter Luschny, May 14 2014
  • Mathematica
    a[ n_] := SeriesCoefficient[ ((1 - x) Sum[ x^k / (1 - (k + 2) x), {k, 0, n}]), {x, 0, n}]; (* Michael Somos, Jul 09 2014 *)
  • PARI
    /* From o.g.f. (Paul D. Hanna, Jul 20 2014) */
    {a(n)=polcoeff( sum(m=0, n, (m+1)!*x^m/(1-x)^(m+1)/prod(k=1, m+1, 1+k*x +x*O(x^n))), n)}
    for(n=0, 25, print1(a(n), ", "))
    
  • PARI
    /* From o.g.f. (Paul D. Hanna, Jul 22 2014) */
    {a(n)=polcoeff( sum(m=0, n, x^m/((1-m*x)*(1-(m+1)*x +x*O(x^n)))), n)}
    for(n=0, 25, print1(a(n), ", "))
  • Sage
    def A074664():
        T = []; n = 0
        while True:
            T.append(1)
            yield T[0]
            for k in (0..n):
                T[k] = (k+1)*T[k] + add(T[j] for j in (k..n))
            n += 1
    a = A074664()
    [next(a) for n in range(25)] # Peter Luschny, May 13 2014
    

Formula

Formal o.g.f.: (1 - x)*( Sum_{n >= 0} x^n/(1 - (n + 2)*x) ). - Peter Bala, Jul 09 2014
O.g.f.: Sum_{n>=0} (n+1)! * x^n/(1-x)^(n+1) / Product_{k=1..n+1} (1 + k*x). - Paul D. Hanna, Jul 20 2014
O.g.f.: Sum_{n>=0} x^n / ( (1 - n*x) * (1 - (n+1)*x) ). - Paul D. Hanna, Jul 22 2014
From Mathew Englander, Feb 28 2021: (Start)
a(n) = A089246(n+2,0) = A242431(n,0).
a(n) = Sum_{m = 1..n+1} Sum_{i = 0..m-1} binomial(m,i) * (n-m+1)^i.
a(n) = 1 + Sum_{i = 0..n} i * (i+1)^(n-i). (End)
a(n) ~ sqrt(2*Pi*n / (w*(1+w))) * (1 + n/w)^(1 + n - n/w), where w = LambertW(exp(1)*n). - Vaclav Kotesovec, Jun 10 2025

A051129 Table T(n,k) = k^n read by upwards antidiagonals (n >= 1, k >= 1).

Original entry on oeis.org

1, 1, 2, 1, 4, 3, 1, 8, 9, 4, 1, 16, 27, 16, 5, 1, 32, 81, 64, 25, 6, 1, 64, 243, 256, 125, 36, 7, 1, 128, 729, 1024, 625, 216, 49, 8, 1, 256, 2187, 4096, 3125, 1296, 343, 64, 9, 1, 512, 6561, 16384, 15625, 7776, 2401, 512, 81, 10, 1, 1024, 19683, 65536, 78125, 46656, 16807, 4096, 729, 100, 11
Offset: 1

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Author

Keywords

Comments

(n-th term) = (n-th term of A002260)^(n-th term of A004736). Both A002260 and A004736 are related to A002024. - Robert A. Stump (bee_ess107(AT)yahoo.com), Aug 29 2002

Examples

			  1   2       3       4       5       6       7
  1   4       9      16      25      36      49
  1   8      27      64     125     216     343
  1  16      81     256     625    1296    2401
  1  32     243    1024    3125    7776   16807
  1  64     729    4096   15625   46656  117649
  1 128    2187   16384   78125  279936  823543
		

Crossrefs

Cf. A051128 (transposed), A003992 (transposed), A004248.
Cf. A002260, A003101 (antidiagonal sums), A000169 (central terms), A003320 (row maxima), A247358 (sorted rows).

Programs

  • Haskell
    a051129 n k = k ^ (n - k)
    a051129_row n = a051129_tabl !! (n-1)
    a051129_tabl = zipWith (zipWith (^)) a002260_tabl $ map reverse a002260_tabl
    -- Reinhard Zumkeller, Sep 14 2014
    
  • Maple
    T:= (n, k)-> k^n:
    seq(seq(T(1+d-k, k), k=1..d), d=1..11);  # Alois P. Heinz, Apr 18 2020
  • Mathematica
    Table[ k^(n-k+1), {n, 1, 12}, {k, 1, n}] // Flatten (* Jean-François Alcover, Nov 30 2012 *)
  • PARI
    b(n) = floor(1/2 + sqrt(2 * n));
    vector(100, n, (n - b(n) * (b(n) - 1) / 2)^(b(n) * (b(n) + 1) / 2 - n + 1)) \\ Altug Alkan, Dec 09 2015

Formula

a(n) = (n - b(n) * (b(n) - 1) / 2)^(b(n) * (b(n) + 1) / 2 - n + 1), where b(n) = [ 1/2 + sqrt(2 * n) ]. (b(n) is the n-th term of A002024.) - Robert A. Stump (bee_ess107(AT)yahoo.com), Aug 29 2002

Extensions

More terms from James Sellers, Dec 11 1999

A287215 Number T(n,k) of set partitions of [n] such that the maximal absolute difference between the least elements of consecutive blocks equals k; triangle T(n,k), n>=0, 0<=k<=max(n-1,0), read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 1, 8, 5, 1, 1, 22, 21, 7, 1, 1, 65, 86, 39, 11, 1, 1, 209, 361, 209, 77, 19, 1, 1, 732, 1584, 1123, 493, 171, 35, 1, 1, 2780, 7315, 6153, 3124, 1293, 413, 67, 1, 1, 11377, 35635, 34723, 20019, 9320, 3709, 1059, 131, 1, 1, 49863, 183080, 202852, 130916, 66992, 30396, 11373, 2837, 259, 1
Offset: 0

Views

Author

Alois P. Heinz, May 21 2017

Keywords

Comments

The maximal absolute difference is assumed to be zero if there are fewer than two blocks.
T(n,k) is defined for all n,k >= 0. The triangle contains only the positive terms. T(n,k) = 0 if k>=n and k>0.

Examples

			T(4,0) = 1: 1234.
T(4,1) = 8: 134|2, 13|24, 14|23, 1|234, 14|2|3, 1|24|3, 1|2|34, 1|2|3|4.
T(4,2) = 5: 124|3, 12|34, 12|3|4, 13|2|4, 1|23|4.
T(4,3) = 1: 123|4.
Triangle T(n,k) begins:
  1;
  1;
  1,   1;
  1,   3,    1;
  1,   8,    5,    1;
  1,  22,   21,    7,   1;
  1,  65,   86,   39,  11,   1;
  1, 209,  361,  209,  77,  19,  1;
  1, 732, 1584, 1123, 493, 171, 35, 1;
		

Crossrefs

Row sums give A000110.
T(2n,n) gives A322884.

Programs

  • Maple
    b:= proc(n, k, m, l) option remember; `if`(n<1, 1,
         `if`(l-n>k, 0, b(n-1, k, m+1, n))+m*b(n-1, k, m, l))
        end:
    A:= (n, k)-> b(n-1, min(k, n-1), 1, n):
    T:= (n, k)-> A(n, k)-`if`(k=0, 0, A(n, k-1)):
    seq(seq(T(n, k), k=0..max(n-1, 0)), n=0..12);
  • Mathematica
    b[n_, k_, m_, l_] := b[n, k, m, l] = If[n < 1, 1, If[l - n > k, 0, b[n - 1, k, m + 1, n]] + m*b[n - 1, k, m, l]];
    A[n_, k_] := b[n - 1, Min[k, n - 1], 1, n];
    T[n_, k_] := A[n, k] - If[k == 0, 0, A[n, k - 1]];
    Table[T[n, k], {n, 0, 12}, {k, 0, Max[n - 1, 0]}] // Flatten (* Jean-François Alcover, May 19 2018, after Alois P. Heinz *)

Formula

T(n,k) = A287216(n,k) - A287216(n,k-1) for k>0, T(n,0) = 1.

A051128 Table T(n,k) = n^k read by upwards antidiagonals (n >= 1, k >= 1).

Original entry on oeis.org

1, 2, 1, 3, 4, 1, 4, 9, 8, 1, 5, 16, 27, 16, 1, 6, 25, 64, 81, 32, 1, 7, 36, 125, 256, 243, 64, 1, 8, 49, 216, 625, 1024, 729, 128, 1, 9, 64, 343, 1296, 3125, 4096, 2187, 256, 1, 10, 81, 512, 2401, 7776, 15625, 16384, 6561, 512, 1, 11, 100, 729, 4096, 16807, 46656, 78125, 65536, 19683, 1024, 1
Offset: 1

Views

Author

Keywords

Comments

Sum of antidiagonals is A003101(n) for n>0. - Alford Arnold, Jan 14 2007

Examples

			Table begins
1,    1,    1,    1,    1, ...
2,    4,    8,   16,   32, ...
3,    9,   27,   81,  243, ...
4,   16,   64,  256, 1024, ...
		

Crossrefs

Programs

Formula

a(n) = A004736(n)^A002260(n) or ((t*t+3*t+4)/2-n)^(n-(t*(t+1)/2)), where t=floor((-1+sqrt(8*n-7))/2). - Boris Putievskiy, Dec 14 2012

Extensions

More terms from James Sellers, Dec 11 1999

A113231 Ascending descending base exponent transform of triangular numbers (A000217).

Original entry on oeis.org

1, 4, 34, 956, 106721, 75818480, 490656737694, 22960404169011552, 7141530219670856270919, 20319415706020976355219258316, 1104797870481014132439711155738991604
Offset: 1

Views

Author

Jonathan Vos Post, Jan 07 2006

Keywords

Comments

A003101 is the ascending descending base exponent transform of natural numbers A000027. The ascending descending base exponent transform applied to the Fibonacci numbers is A113122; applied to the tribonacci numbers is A113153; applied to the Lucas numbers is A113154. Since the parity of the triangular numbers cycles odd, odd, even, even; the parity of this sequence cycles odd, even, even, even. The smallest prime in this sequence is a(5) = 127601. What is the next prime? What is the first triangular value?

Examples

			a(1) = 1 because T(1)^T(1) = 1^1 = 1.
a(2) = 4 because T(1)^T(2) + T(2)^T(1) = 1^3 + 3^1 = 4.
a(3) = 34 = 1^6 + 3^3 + 6^1.
a(4) = 956 = 1^10 + 3^6 + 6^3 + 10^1.
a(5) = 106721 = 1^15 + 3^10 + 6^6 + 10^3 + 15^1.
a(6) = 75818480 = 1^21 + 3^15 + 6^10 + 10^6 + 15^3 + 21^1.
a(7) = 490656737694 = 1^28 + 3^21 + 6^15 + 10^10 + 15^6 + 21^3 + 28^1.
a(8) = 22960404169011552 = 1^36 + 3^28 + 6^21 + 10^15 + 15^10 + 21^6 + 28^3 + 36^1.
a(9) = 7141530219670856270919 = 1^45 + 3^36 + 6^28 + 10^21 + 15^15 + 21^10 + 28^6 + 36^3 + 45^1.
		

Crossrefs

Programs

  • Mathematica
    A000217[n_] := Binomial[n + 1, 2]; Table[Sum[A000217[k]^(A000217[n - k + 1]), {k, 1, n}], {n, 1, 10}] (* G. C. Greubel, May 18 2017 *)
  • PARI
    for(n=1,10, print1(sum(k=1,n, (binomial(k+1,2))^(binomial(n-k+2,2))), ", ")) \\ G. C. Greubel, May 18 2017

Formula

a(n) = Sum_{i=1..n} (T(i))^(T(n-i+1)), where T(n) are the triangle numbers.
a(n) = Sum_{i=1..n} ((i*(i+1)/2))^((n-i+1)*(n-i+2)/2).
a(n) = Sum_{i=1..n} (A000217(i))^(A000217(n-i+1)).
log(a(n)) ~ n^2 * (-1 + 2*LambertW(2^(-3/2)*exp(1/2)*n))^3 / (8*LambertW(2^(-3/2)*exp(1/2)*n)^2). - Vaclav Kotesovec, Jun 07 2025

A113257 Ascending descending base exponent transform of squares (A000290).

Original entry on oeis.org

1, 5, 266, 268722, 4682453347, 2978988815561863, 722638800922610642480852, 22529984108212742763058965679103268, 57286470055793196612331429228839529219232484069
Offset: 1

Views

Author

Jonathan Vos Post, Jan 07 2006

Keywords

Comments

A003101 is the ascending descending base exponent transform of natural numbers A000027. The ascending descending base exponent transform applied to the Fibonacci numbers is A113122; applied to the tribonacci numbers is A113153; applied to the Lucas numbers is A113154. The smallest prime in this sequence is a(2) = 5. What is the next prime? What is the first square value after 1?

Examples

			a(1) = 1 because (1^2)^(1^2) = 1^1 = 1.
a(2) = 5 because (1^2)^(4^1) + (4^1)^(1^4) = 1^4 + 4^1 = 5.
a(3) = 266 = 1^9 + 4^4 + 9^1.
a(4) = 268722 = 1^16 + 4^9 + 9^4 + 16^1.
a(5) = 4682453347 = 1^25 + 4^16 + 9^9 + 16^4 + 25^1.
a(6) = 2978988815561863 = 1^36 + 4^25 + 9^16 + 16^9 + 25^4 + 36^1.
a(7) = 722638800922610642480852 = 1^49 + 4^36 + 9^25 + 16^16 + 25^9 + 36^4 + 49^1.
a(8) = 22529984108212742763058965679103268 = 1^64 + 4^49 + 9^36 + 16^25 + 25^16 + 36^9 + 49^4 + 64^1.
a(9) = 57286470055793196612331429228839529219232484069 = 1^81 + 4^64 + 9^49 + 16^36 + 25^25 + 36^16 + 49^9 + 64^4 + 81^1.
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[(k^2)^((n - k + 1)^2), {k, 1, n}], {n, 1, 10}] (* G. C. Greubel, May 18 2017 *)
  • PARI
    for(n=1,10, print1(sum(k=1,n, (k^2)^((n-k+1)^2) ), ", ")) \\ G. C. Greubel, May 18 2017

Formula

a(n) = Sum_{i=1..n} (i^2)^((n-i+1)^2).
a(n) = Sum_{i=1..n} (A000290(i))^(A000290(n-i+1)).
log(a(n)) ~ n^2 * (-1 + 2*LambertW(exp(1/2)*n/2))^3 / (4*LambertW(exp(1/2)*n/2)^2). - Vaclav Kotesovec, Jun 07 2025

Extensions

a(4) and a(5) corrected by Giovanni Resta, Jun 13 2016

A113271 Ascending descending base exponent transform of 2^n.

Original entry on oeis.org

1, 3, 9, 41, 593, 135457, 8606778433, 36893769626691833985, 680564733921105089459460297630318346497, 231584178474632390853419071752762496470716041121409734167406717963826481922561
Offset: 0

Views

Author

Jonathan Vos Post, Jan 07 2006

Keywords

Comments

A003101 is the ascending descending base exponent transform of natural numbers A000027. The ascending descending base exponent transform applied to the Fibonacci numbers is A113122; applied to the tribonacci numbers is A113153; applied to the Lucas numbers is A113154. The smallest primes in this (always odd) sequence are a(1) = 3, a(3) = 41 and a(5) = 543. What is the next prime?

Examples

			a(0) = 1 because (2^0)^(2^0) = 1^1 = 1.
a(1) = 3 = (2^0)^(2^1) + (2^1)^(2^0) = 1^2 + 2^1.
a(2) = 9 = (2^0)^(2^2) + (2^1)^(2^1) + (2^2)^(2^0) = 1^4 + 2^2 + 4^1.
a(3) = 41 = 1^8 + 2^4 + 4^2 + 8^1.
a(4) = 593 = 1^16 + 2^8 + 4^4 + 8^2 + 16^1
a(5) = 135457 = 1^32 + 2^16 + 4^8 + 8^4 + 16^2 + 32^1.
a(6) = 8606778433 = 1^64 + 2^32 + 4^16 + 8^8 + 16^4 + 32^2 + 64^1.
a(7) = 36893769626691833985 = 1^128 + 2^64 + 4^32 + 8^16 + 16^8 + 32^4 + 64^2 + 128^1.
a(8) = 680564733921105089459460297630318346497 = 1^256 + 2^128 + 4^64 + 8^32 + 16^16 + 32^8 + 64^4 + 128^2 + 256^1.
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[(2^k)^(2^(n - k)), {k, 0, n}], {n, 0, 10}] (* G. C. Greubel, May 19 2017 *)
  • PARI
    for(n=0,5, print1(sum(k=0,n, (2^k)^(2^(n-k))), ", ")) \\ G. C. Greubel, May 19 2017

Formula

a(n) = Sum_{i=0..n} (2^i)^(2^(n-i)).
a(n) = Sum_{i=0..n} (2^(n-i))^(2^i).
a(n) = Sum_{i=0..n} (A000079(i))^(A000079(n-i)).
a(n) ~ 2^(2^(n-1) + 1). - Vaclav Kotesovec, Jun 07 2025

Extensions

a(4) corrected by Giovanni Resta, Jun 13 2016
Formulas corrected by G. C. Greubel, May 19 2017

A038125 a(n) = Sum_{k=0..n} (k-n)^k.

Original entry on oeis.org

1, 1, 0, 0, 1, -1, 0, 6, -19, 29, 48, -524, 2057, -3901, -9632, 129034, -664363, 1837905, 2388688, -67004696, 478198545, -1994889945, 1669470784, 56929813934, -615188040195, 3794477505573, -12028579019536, -50780206473220
Offset: 0

Views

Author

Jim Ferry (jferry(AT)alum.mit.edu)

Keywords

Examples

			0^0 = 1,
1^0 - 0^1 = 1,
2^0 - 1^1 + 0^2 = 0,
3^0 - 2^1 + 1^2 - 0^3 = 0,
...
		

Crossrefs

Programs

  • Mathematica
    Prepend[ Table[ Sum[ (k-n)^k, {k, 0, n} ], {n, 30} ], 1 ]
  • PARI
    my(N=40, x='x+O('x^N)); Vec(sum(k=0, N, x^k/(1+k*x))) \\ Seiichi Manyama, Dec 02 2021
    
  • PARI
    a(n) = sum(k=0, n, (k-n)^k); \\ Michel Marcus, Dec 03 2021

Formula

G.f.: 1+ sum(k>=0, x^(k+1)/(1+x^(k+1)) ) = 1/Q(0), where Q(k) = 1 - x + x^2*(k+1)/(1 + (k+1)*x/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Jan 10 2014

A113258 Ascending descending base exponent transform of factorials.

Original entry on oeis.org

1, 3, 11, 125, 16824569, 1329227995784915877642188398793079569
Offset: 1

Views

Author

Jonathan Vos Post, Jan 07 2006

Keywords

Comments

A003101 is the ascending descending base exponent transform of natural numbers A000027. The ascending descending base exponent transform applied to the Fibonacci numbers is A113122; applied to the tribonacci numbers is A113153; applied to the Lucas numbers is A113154. The smallest primes in this (always odd) sequence are a(2) = 3 and a(3) = 11. What is the next prime? Is there a nontrivial power after a(4) = 5^3?

Examples

			a(1) = 1 because (1!)^(1!) = 1^1 = 1.
a(2) = 3 because (1!)^(2!) + (2!)^(1!) = 1 + 2 = 3.
a(3) = 11 = (1!)^(3!) + (2!)^(2!) + (3!)^(1!) = 1^6 + 2^2 + 6^1 = 11.
a(4) = 125 = (1!)^(4!) + (2!)^(3!) + (3!)^(2!) + (4!)^(1!).
a(6) = 1329227995784915877642188398793079569 = 1^720 + 2^120 + 6^24 + 24^6 + 120^2 + 720^1.
a(7) = 1!^7! + 2!^6! + 3!^5! + 4!^4! + 5!^3! + 6!^2! + 7!^1! has 217 digits.
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[((k)!)^(n - k + 1)!, {k, 1, n}], {n,1,5}] (* G. C. Greubel, May 18 2017 *)
  • PARI
    for(n=1,5, print1(sum(k=1,n, (k!)^((n-k+1)!)), ", ")) \\ G. C. Greubel, May 18 2017

Formula

a(n) = Sum_{i = 1..n} (i!)^((n-i+1)!).
a(n) = Sum_{i = 1..n} (n-i+1)!^i!.
a(n) = Sum_{i = 1..n} (A000142(i))^(A000142(n-i+1)).
a(n) ~ 2^((n-1)!). - Vaclav Kotesovec, Jun 08 2025
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