A244589 E.g.f. A(x) satisfies the property that the sum of the coefficients of x^k, k=0..n, in A(x)^n equals (n+1)^n.
1, 1, 5, 67, 1937, 98791, 7744549, 857382695, 126889656641, 24157912257775, 5749369223697701, 1672527291075462559, 584038879457972531185, 241150002566590866157943, 116245385996298375640197893, 64707252902905394310560934391, 41198982747438307655532993553409
Offset: 0
Keywords
Examples
E.g.f.: A(x) = 1 + x + 5*x^2/2! + 67*x^3/3! + 1937*x^4/4! + 98791*x^5/5! +... where ILLUSTRATION OF INITIAL TERMS. If we form an array of coefficients of x^k/k! in A(x)^n, n>=0, like so: A^0: [1],0, 0, 0, 0, 0, 0, 0, ...; A^1: [1, 1], 5, 67, 1937, 98791, 7744549, 857382695, ...; A^2: [1, 2, 12], 164, 4560, 223652, 17054920, 1853019716, ...; A^3: [1, 3, 21, 297], 8049, 380853, 28237293, 3008400909, ...; A^4: [1, 4, 32, 472, 12608], 577864, 41657008, 4348646600, ...; A^5: [1, 5, 45, 695, 18465, 823475], 57747565, 5903103995, ...; A^6: [1, 6, 60, 972, 25872, 1127916, 77020344], 7706019180, ...; A^7: [1, 7, 77, 1309, 35105, 1502977, 100075045, 9797289761], ...; ... then we can illustrate how the sum of the coefficients of x^k, k=0..n, in A(x)^n (shown above in brackets) equals (n+1)^n: 1^0 = 1; 2^1 = 1 + 1 = 2; 3^2 = 1 + 2 + 12/2! = 9; 4^3 = 1 + 3 + 21/2! + 297/3! = 64; 5^4 = 1 + 4 + 32/2! + 472/3! + 12608/4! = 625; 6^5 = 1 + 5 + 45/2! + 695/3! + 18465/4! + 823475/5! = 7776; 7^6 = 1 + 6 + 60/2! + 972/3! + 25872/4! + 1127916/5! + 77020344/6! = 117649; ...
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 0..150
Programs
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PARI
/* By Definition (slow): */ {a(n)=if(n==0, 1, n!*((n+1)^n - sum(k=0, n, polcoeff(sum(j=0, min(k, n-1), a(j)*x^j/j!)^n + x*O(x^k), k)))/n)} for(n=0, 20, print1(a(n), ", ")) -
PARI
/* Faster, using series reversion: */ {a(n)=local(B=sum(k=0, n+1, (k+1)^k*x^k)+x^3*O(x^n), G=1+x*O(x^n)); for(i=1, n, G = 1 + intformal( (B-1)*G/x - B*G^2)); n!*polcoeff(x/serreverse(x*G), n)} for(n=0, 30, print1(a(n), ", "))
Formula
E.g.f. A(x) satisfies: Sum_{k=0..n} [x^k] A(x)^n = (n+1)^n.
a(n) ~ exp(1-exp(-1)) * n! * n^(n-1). - Vaclav Kotesovec, Jul 03 2014