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 A233319 E.g.f. A(x) satisfies: A'(x) = A(x*A'(x))^3. 2
 1, 1, 3, 33, 726, 25236, 1229328, 78167484, 6193726506, 592068123054, 66673324219176, 8685890001564984, 1290531658541292252, 216188985806157611520, 40449991773179254230432, 8386998677130790903212000, 1914263814914709029067344724, 478208364783447353623777136772 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS CONJECTURES: a(n) == 0 (mod 2) for n>=4. a(n) == 0 (mod 2^2) for n>=10. a(n) == 0 (mod 2^3) for n>=18. a(n) == 0 (mod 2^k) for n>=(8*k-6), k>=2. a(n) == 0 (mod 3) for n>=2. a(n) == 0 (mod 3^2) for n>=5. a(n) == 0 (mod 3^3) for n>=7. a(n) == 0 (mod 3^4) for n>=10. a(n) == 0 (mod 3^k) for n>=(3*k-2), k>=3. a(n) == 0 (mod 13) for n>=13. LINKS Paul D. Hanna, Table of n, a(n) for n = 0..100 FORMULA E.g.f. A(x) satisfies: A(x)^3 = A'(x/A(x)^3). E.g.f. A(x) satisfies: A(x) = ( x / Series_Reversion( x*A'(x) ) )^(1/3). a(n) = [x^(n-1)/(n-1)!] A(x)^(3*n)/n for n>=1. EXAMPLE E.g.f.: A(x) = 1 + x + 3*x^2/2! + 33*x^3/3! + 726*x^4/4! + 25236*x^5/5! +... such that A(x*A'(x))^3 = A'(x) = 1 + 3*x + 33*x^2/2! + 726*x^3/3! + 25236*x^4/4! +... A(x*A'(x)) = (A'(x))^(1/3) = 1 + x + 9*x^2/2! + 186*x^3/3! + 6330*x^4/4! + 306846*x^5/5! + 19560006*x^6/6! + 1559472498*x^7/7! +... To illustrate a(n) = [x^(n-1)/(n-1)!] A(x)^(3*n)/n, create a table of coefficients of x^k/k!, k>=0, in A(x)^(3*n) like so: A^3: [1, 3, 15, 159, 3240, 106218, 4961250, 305900982, ...]; A^6: [1, 6, 48, 588, 11646, 357336, 15709968, 923153004, ...]; A^9: [1, 9, 99, 1449, 30078, 899964, 37750104, 2118453588, ...]; A^12:[1, 12, 168, 2904, 65340, 1977912, 80833248, 4365682056, ...]; A^15:[1, 15, 255, 5115, 126180, 3961350, 161145630, 8476536330, ...]; A^18:[1, 18, 360, 8244, 223290, 7375968, 304020000, 15786282132, ...]; A^21:[1, 21, 483, 12453, 369306, 12932136, 547172388, 28405637064, ...]; A^24:[1, 24, 624, 17904, 578808, 21554064, 944463744, 49549812048, ...]; ... then the diagonal in the above table generates this sequence shift left: [1/1, 6/2, 99/3, 2904/4, 126180/5, 7375968/6, 547172388/7, 49549812048/8, ...]. SUMS OF TERM RESIDUES MODULO 2^n. Given a(k) == 0 (mod 2^n) for k>=(8*n-6) for n>=2, then it is interesting to consider the sums of the residues of all terms modulo 2^n for n>=1. Let b(n) = Sum_{k>=0} a(k) (mod 2^n) for n>=1, then the sequence {b(n)} begins: [4, 12, 40, 112, 336, 848, 2128, 5584, 13776, 29648, 64464, 136144, 316368, ...]. SUMS OF TERM RESIDUES MODULO 3^n. Given a(k) == 0 (mod 3^n) for k>=(3*n-2) for n>=3, then it is interesting to consider the sums of the residues of all terms modulo 3^n for n>=1. Let c(n) = Sum_{k>=0} a(k) (mod 3^n) for n>=1, then the sequence {c(n)} begins: [2, 17, 71, 368, 1340, 4985, 13733, 59660, 217124, 689516, 2520035, 6594416, 18286118, 72493100, 206416232, 722976884, 2617032608, 8170059617, 25603981622, 93015146708, 256894013555, 832213439720, 2338504300952, 6292517811686, 24650437682951, 71251311202316, 249181919185346, 729594560739527, ...]. PROG (PARI) {a(n)=local(A=1+x); for(i=1, n, A=1+intformal(subst(A^3, x, x*A' +x*O(x^n)))); n!*polcoeff(A, n)} for(n=0, 25, print1(a(n), ", ")) (PARI) {a(n)=local(A=1+x); for(i=1, n, A=1+intformal(1/x*serreverse(x/A^3 +x*O(x^n)))); n!*polcoeff(A, n)} for(n=0, 25, print1(a(n), ", ")) CROSSREFS Cf. A232686, A231619, A231866, A231899, A232694, A232695, A232696. Sequence in context: A091462 A340971 A326328 * A003715 A247030 A009690 Adjacent sequences: A233316 A233317 A233318 * A233320 A233321 A233322 KEYWORD nonn AUTHOR Paul D. Hanna, Dec 07 2013 STATUS approved

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Last modified December 5 05:37 EST 2023. Contains 367575 sequences. (Running on oeis4.)