A001923
a(n) = Sum_{k=1..n} k^k.
Original entry on oeis.org
0, 1, 5, 32, 288, 3413, 50069, 873612, 17650828, 405071317, 10405071317, 295716741928, 9211817190184, 312086923782437, 11424093749340453, 449317984130199828, 18896062057839751444, 846136323944176515621
Offset: 0
- József Sándor, Dragoslav S. Mitrinovic, Borislav Crstici, Handbook of Number Theory I, Springer Science & Business Media, 2005, p. 308.
- N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
- N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
- T. D. Noe, Table of n, a(n) for n=0..100
- Mohammad K. Azarian, On the hyperfactorial function, hypertriangular function and the discriminants of certain polynomials, Int. J. Pure Appl. Math., Vol. 36, No. 2 (2007), pp. 251-257.
- Andrew Cusumano, Problem H-656, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 45, No. 2 (2007), p. 187; A Sequence Tending To e, Solution to Problem H-656, ibid., Vol. 46-47, No. 3 (2008/2009), pp. 285-287.
- G. W. Wishard (proposer) and F. Underwood (solution), Problem 4155: Bound for a Finite Sum, Amer. Math. Monthly, Vol. 53, No. 8 (1946), pp. 471-473.
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a001923 n = a001923_list !! n
a001923_list = scanl (+) 0 $ tail a000312_list
-- Reinhard Zumkeller, Jul 11 2014
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Accumulate[Join[{0},Table[k^k,{k,20}]]] (* Harvey P. Dale, Feb 11 2015 *)
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for(n=1,20,print1(sum(x=1,n,x^x), ", ")) \\ Jorge Coveiro, Dec 24 2004
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# generates initial segment of sequence
from itertools import accumulate
def f(k): return 0 if k == 0 else k**k
def aupton(nn): return list(accumulate(f(k) for k in range(nn+1)))
print(aupton(17)) # Michael S. Branicky, Feb 12 2022
A209424
Triangle defined by g.f.: A(x,y) = exp( Sum_{n>=1} x^n/n * Sum_{k=0..n} binomial(n, k)^n * y^k ), as read by rows.
Original entry on oeis.org
1, 1, 1, 1, 3, 1, 1, 12, 12, 1, 1, 76, 347, 76, 1, 1, 701, 20429, 20429, 701, 1, 1, 8477, 1919660, 10707908, 1919660, 8477, 1, 1, 126126, 259227625, 9203978774, 9203978774, 259227625, 126126, 1, 1, 2223278, 47484618291, 12099129236936, 72078431500368
Offset: 0
This triangle begins:
1;
1, 1;
1, 3, 1;
1, 12, 12, 1;
1, 76, 347, 76, 1;
1, 701, 20429, 20429, 701, 1;
1, 8477, 1919660, 10707908, 1919660, 8477, 1;
1, 126126, 259227625, 9203978774, 9203978774, 259227625, 126126, 1;
1, 2223278, 47484618291, 12099129236936, 72078431500368, 12099129236936, 47484618291, 2223278, 1; ...
G.f.: A(x,y) = 1 + (1+y)*x + (1+3*y+y^2)*x^2 + (1+12*y+12*y^2+y^3)*x^3 + (1+76*y+20429*y^2+76*y^3+y^4)*x^4 +...
The logarithm of the g.f. equals the series:
log(A(x,y)) = (1 + y)*x
+ (1 + 2^2*y + y^2)*x^2/2
+ (1 + 3^3*y + 3^3*y^2 + y^3)*x^3/3
+ (1 + 4^4*y + 6^4*y^2 + 4^4*y^3 + y^4)*x^4/4
+ (1 + 5^5*y + 10^5*y^2 + 10^5*y^3 + 5^5*y^4 + y^5)*x^5/5 +...
in which the coefficients are found in triangle A209427.
-
{T(n,k)=polcoeff(polcoeff(exp(sum(m=1,n,x^m/m*sum(k=0,m,binomial(m,k)^m*y^k))+x*O(x^n)),n,x),k,y)}
for(n=0,10,for(k=0,n,print1(T(n,k),", "));print(""))
A345094
a(n) = Sum_{k=1..n} floor(n/k)^(floor(n/k) - 1).
Original entry on oeis.org
1, 3, 11, 68, 630, 7790, 117664, 2097224, 43046801, 1000000643, 25937425245, 743008378547, 23298085130341, 793714773371879, 29192926025508929, 1152921504608944840, 48661191875668966346, 2185911559738739586562, 104127350297911284587436
Offset: 1
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a[n_] := Sum[Floor[n/k]^(Floor[n/k] - 1), {k, 1, n}]; Array[a, 20] (* Amiram Eldar, Jun 08 2021 *)
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a(n) = sum(k=1, n, (n\k)^(n\k-1));
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my(N=20, x='x+O('x^N)); Vec(sum(j=1, N, (1-x^j)*sum(k=1, N, k^(k-1)*x^(j*k)))/(1-x))
A355950
a(n) = Sum_{k=1..n} k^(k-1) * floor(n/k).
Original entry on oeis.org
1, 4, 14, 81, 707, 8495, 126145, 2223364, 45270095, 1045270723, 26982695325, 769991073865, 24068076196347, 817782849568143, 30010708874959403, 1182932213483903598, 49844124089150772080, 2235755683827890358557, 106363105981739131891399
Offset: 1
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a(n) = sum(k=1, n, n\k*k^(k-1));
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a(n) = sum(k=1, n, sumdiv(k, d, d^(d-1)));
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my(N=20, x='x+O('x^N)); Vec(sum(k=1, N, k^(k-1)*x^k/(1-x^k))/(1-x))
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def A355950(n): return n*(1+n**(n-2))+sum(k**(k-1)*(n//k) for k in range(2,n)) if n>1 else 1 # Chai Wah Wu, Jul 21 2022
A228899
Triangle defined by g.f. A(x,y) = exp( Sum_{n>=1} x^n/n * Sum_{k=0..n} binomial(n, k)^(k+1) * y^k ), as read by rows.
Original entry on oeis.org
1, 1, 1, 1, 3, 1, 1, 6, 12, 1, 1, 10, 71, 76, 1, 1, 15, 281, 2153, 701, 1, 1, 21, 861, 29166, 129509, 8477, 1, 1, 28, 2212, 244725, 7664343, 12391414, 126126, 1, 1, 36, 4998, 1477391, 218030412, 3875325345, 1699148352, 2223278, 1, 1, 45, 10242, 7017577, 3748460115, 448713017405, 3284369541969, 315158247170, 45269999, 1
Offset: 0
This triangle begins:
1;
1, 1;
1, 3, 1;
1, 6, 12, 1;
1, 10, 71, 76, 1;
1, 15, 281, 2153, 701, 1;
1, 21, 861, 29166, 129509, 8477, 1;
1, 28, 2212, 244725, 7664343, 12391414, 126126, 1;
1, 36, 4998, 1477391, 218030412, 3875325345, 1699148352, 2223278, 1;
1, 45, 10242, 7017577, 3748460115, 448713017405, 3284369541969, 315158247170, 45269999, 1; ...
...
G.f.: A(x,y) = 1 + (1+y)*x + (1+3*y+y^2)*x^2 + (1+6*y+12*y^2+y^3)*x^3 + (1+10*y+71*y^2+76*y^3+y^4)*x^4 + (1+15*y+281*y^2+2153*y^3+701*y^4+y^5)*x^5 +...
The logarithm of the g.f. equals the series:
log(A(x)) = (1 + x)*x
+ (1 + 2^2*x + x^2)*x^2/2
+ (1+ 3^2*y + 3^3*y^2 + y^3)*x^3/3
+ (1+ 4^2*y + 6^3*y^2 + 4^4*y^3 + x^4)*x^4/4
+ (1+ 5^2*y + 10^3*y^2 + 10^4*y^3 + 5^5*y^4 + y^5)*x^5/5
+ (1+ 6^2*y + 15^3*y^2 + 20^4*y^3 + 15^5*y^4 + 6^6*y^5 + y^6)*x^6/6 +...
in which the coefficients form A219207(n,k) = binomial(n, k)^(k+1).
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{T(n, k)=polcoeff(polcoeff(exp(sum(m=1, n, x^m/m*sum(j=0, m, binomial(m, j)^(j+1)*y^j))+x*O(x^n)), n, x), k, y)}
for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print(""))
A232821
a(n) = n^(n-1) - Sum_{k=1..n-1} k^(k-1).
Original entry on oeis.org
1, 1, 6, 52, 549, 7075, 109172, 1971026, 40823443, 954730001, 24892154602, 716025676088, 22528094057193, 769646697066375, 28375143175948712, 1122910795732014438, 47478259662185188967, 2136067435649547983973, 101891594614083396452878
Offset: 1
6^5 - 5^4 - 4^3 - 3^2 - 2^1 - 1^0 = 7075 so a(6) = 7075.
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a[n_] := n^(n - 1) - Sum[i^(i - 1), {i, 1, n - 1}]; Table[a[n], {n, 20}] (* Carlos Eduardo Olivieri, May 29 2015 *)
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vector(20,n,n^(n-1)-sum(i=1,n-1,i^(i-1))) \\ Derek Orr, Apr 05 2015
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def sub(n):
num = n**(n-1)
for i in range(0, n-1):
num -= (i+1)**i
return num
n = 1
while n < 100:
print(sub(n), end=', ')
n += 1
A128884
Sum of all matrix elements of n X n Vandermonde matrix of numbers 1,2,...,n, i.e., the matrix A with A[i,j] = i^(j-1), 1 <= i <= n, 1 <= j <= n.
Original entry on oeis.org
1, 5, 23, 144, 1279, 15035, 219463, 3816512, 76928685, 1762344781, 45207853767, 1283438430208, 39944988007339, 1352308628695895, 49471532968242991, 1944732944768690432, 81748776383970349721, 3659142661552743151353
Offset: 1
Cf.
A060946 = Trace of Vandermonde matrix of numbers 1, 2, ..., n.
Cf.
A000178 = Superfactorials: product of first n factorials.
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Table[ n + Sum[ (i^n-1)/(i-1), {i,2,n} ], {n,1,25} ]
A276455
Primes of the form Sum_{k=1..n} k^(k-1).
Original entry on oeis.org
3, 701, 45269999
Offset: 1
3 is in the sequence because 3 is prime and 3 = 2^1 + 1^0.
701 is in the sequence because 701 is prime and 701 = 5^4 + 4^3 + 3^2 + 2^1 + 1^0.
45269999 is in the sequence because 45269999 is prime and 45269999 = 9^8 + 8^7 + 7^6 + 6^5 + 5^4 + 4^3 + 3^2 + 2^1 + 1^0.
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Select[Accumulate[Table[n^(n-1),{n,100}]],PrimeQ] (* Harvey P. Dale, Apr 13 2020 *)
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sum = 0
seq = []
max_n = 2500
for n in range(1, max_n+1):
sum += n^(n-1)
if is_prime(sum):
seq.append(n)
print(seq)
A349928
a(n) = Sum_{k=0..n} (k+n)^k.
Original entry on oeis.org
1, 3, 20, 246, 4481, 107129, 3157836, 110504876, 4473749677, 205615442135, 10574135574388, 601527803412298, 37500537926181449, 2542321872054610333, 186209553386691383388, 14653121207168215024624, 1232879877057607865696085, 110444572988776439826640683
Offset: 0
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a[0] = 1; a[n_] := Sum[(k + n)^k, {k, 0, n}]; Array[a, 18, 0] (* Amiram Eldar, Dec 05 2021 *)
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a(n) = sum(k=0, n, (k+n)^k);
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