A206295 (1/6)*A007188(n).
1, 15, 7875, 11142140625, 467812950752090302734375, 202822649430450649753225796950422853099588897705078125
Offset: 1
Keywords
Crossrefs
Cf. A007188.
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.
Array T(n, k) begins: n\k| 1 2 3 4 5 6 7 8 9 10 ---+------------------------------------------------ 1| 1 1 1 1 1 1 1 1 1 1 -> A000012 2| 1 2 3 4 5 6 7 8 9 10 -> A000027 3| 1 3 5 9 7 15 11 27 25 21 -> A003961 4| 1 4 9 16 25 36 49 64 81 100 -> A000290 5| 1 5 7 25 11 35 13 125 49 55 -> A357852 6| 1 6 15 36 35 90 77 216 225 210 -> A191002 7| 1 7 11 49 13 77 17 343 121 91 8| 1 8 27 64 125 216 343 512 729 1000 -> A000578 9| 1 9 25 81 49 225 121 729 625 441 10| 1 10 21 100 55 210 91 1000 441 550 From _Peter Munn_, Jun 24 2021: (Start) The encoding, n, of polynomials, f(n), that is used for the table is further described in A206284. Examples of encoded polynomials: n f(n) n f(n) 1 0 16 4 2 1 17 x^6 3 x 21 x^3 + x 4 2 25 2x^2 5 x^2 27 3x 6 x + 1 35 x^3 + x^2 7 x^3 36 2x + 2 8 3 49 2x^3 9 2x 55 x^4 + x^2 10 x^2 + 1 64 6 11 x^4 77 x^4 + x^3 12 x + 2 81 4x 13 x^5 90 x^2 + 2x + 1 15 x^2 + x 91 x^5 + x^3 (End)
T(n,k) = my (f=factor(n), p=apply(primepi, f[, 1]~), g=factor(k), q=apply(primepi, g[, 1]~)); prod (i=1, #p, prod(j=1, #q, prime(p[i]+q[j]-1)^(f[i, 2]*g[j, 2])))
n a(n) prime factorization Fibonacci polynomial ------------------------------------------------------------ 0 1 (empty) F_0(x) = 0 1 2 p_1 F_1(x) = 1 2 3 p_2 F_2(x) = x 3 10 p_3 * p_1 F_3(x) = x^2 + 1 4 63 p_4 * p_2^2 F_4(x) = x^3 + 2x 5 2750 p_5 * p_3^3 * p_1 F_5(x) = x^4 + 3x^2 + 1 6 842751 p_6 * p_4^4 * p_2^3 F_6(x) = x^5 + 4x^3 + 3x
c[n_] := CoefficientList[Fibonacci[n, x], x] f[n_] := Product[Prime[k]^c[n][[k]], {k, 1, Length[c[n]]}] Table[f[n], {n, 1, 11}] (* A206296 *)
from functools import reduce from sympy import factorint, prime, primepi from operator import mul def a003961(n): F=factorint(n) return 1 if n==1 else reduce(mul, [prime(primepi(i) + 1)**F[i] for i in F]) l=[1, 2] for n in range(2, 11): l.append(a003961(l[n - 1])*l[n - 2]) print(l) # Indranil Ghosh, Jun 21 2017
T(4,3) = T(3,2)*T(4,2) = 15*35 = 525. Rows start 2; 3, 6; 5, 15, 90; 7, 35, 525, 47250; ... From _Antti Karttunen_, Sep 18 2016: (Start) Alternatively, this table can be viewed as a square array. Then the top left 5x4 corner looks as: 2, 3, 5, 7, 11 6, 15, 35, 77, 143 90, 525, 2695, 11011, 31603 47250, 1414875, 29674645, 347980633, 2255916949 (End)
T[n_, 1] := Prime[n]; T[n_, k_] := T[n, k] = T[n - 1, k - 1]*T[n, k - 1]; Table[T[n, k], {n, 1, 7}, {k, 1, n}] // Flatten (* Jean-François Alcover, Nov 13 2017 *)
(define (A066117 n) (A066117bi (A002260 n) (A004736 n))) ;; Compute as a square array, with row >= 1, col >= 1: (define (A066117bi row col) (if (= 1 row) (A000040 col) (* (A066117bi (- row 1) col) (A066117bi (- row 1) (+ col 1))))) ;; With alternative recurrence: (define (A066117bi row col) (if (= 1 col) (A007188 (- row 1)) (A003961 (A066117bi row (- col 1))))) ;; Antti Karttunen, Sep 18 2016
a(0) = 2^T(0,0) = 2^1 = 2. a(1) = 2^T(1,0) * 3^T(1,1) = 2^1 * 3^1 = 6. a(2) = 2^T(2,0) * 3^T(2,1) * 5^T(2,2) = 2^1 * 3^0 * 5^1 = 10. a(3) = 2^T(3,0) * 3^T(3,1) * 5^T(3,2) * 7^T(3,3) = 2^1 * 3^1 * 5^1 * 7^1 = 210. a(4) = 2^1 * 3^0 * 5^0 * 7^0 * 11^1 = 22. a(5) = 2^1 * 3^1 * 5^0 * 7^0 * 11^1 * 13^1 = 858. a(6) = 2^1 * 3^0 * 5^1 * 7^0 * 11^1 * 13^0 * 17^1 = 1870. a(7) = 2^1 * 3^1 * 5^1 * 7^1 * 11^1 * 13^1 * 17^1 * 19^1 = 9699690. a(8) = 2^1 * 3^0 * 5^0 * 7^0 * 11^0 * 13^0 * 17^0 * 19^0 * 23^1 = 46. a(9) = 2^1 * 3^1 * 5^0 * 7^0 * 11^0 * 13^0 * 17^0 * 19^0 * 23^1 * 29^1 = 4002. a(10) = 2^1 * 3^0 * 5^1 * 7^0 * 11^0 * 13^0 * 17^0 * 19^0 * 23^1 * 29^0 * 31^1 = 7130. a(11) = 2^1 * 3^1 * 5^1 * 7^1 * 11^0 * 13^0 * 17^0 * 19^0 * 23^1 * 29^1 * 31^1 * 37^1 = 160660290. a(12) = 2^1 * 3^0 * 5^0 * 7^0 * 11^1 * 13^0 * 17^0 * 19^0 * 23^1 * 29^0 * 31^0 * 37^0 * 41^1 = 20746. From _N. J. A. Sloane_, Feb 28 2015: (Start) Factorizations of initial terms, from Cobeli-Zaharescu paper: 2 = 2 6 = 2*3 10 = 2*5 210 = 2*3*5*7 22 = 2*11 858 = 2*3*11*13 1870 = 2*5*11*17 9699690 = 2*3*5*7*11*13*17*19 46 = 2*23 4002 = 2*3*23*29 7130 = 2*5*23*31 160660290 = 2*3*5*7*23*29*31*37 20746 = 2*11*23*41 1008940218 = 2*3*11*13*23*29*41*43 2569288370 = 2*5*11*17*23*31*41*47 32589158477190044730 = 2*3*5*7*11*13*17*19*23*29*31*37*41*43*47*53 ... (End) From _Jon E. Schoenfield_, Jun 09 2019: (Start) n | Factorization of a(n) ---+----------------------------------------------- 0 | 2 1 | 2* 3 2 | 2 * 5 3 | 2* 3* 5* 7 4 | 2 *11 5 | 2* 3 *11*13 6 | 2 * 5 *11 *17 7 | 2* 3* 5* 7*11*13*17*19 8 | 2 *23 9 | 2* 3 *23*29 10 | 2 * 5 *23 *31 11 | 2* 3* 5* 7 *23*29*31*37 12 | 2 *11 *23 *41 13 | 2* 3 *11*13 *23*29 *41*43 14 | 2 * 5 *11 *17 *23 *31 *41 *47 15 | 2* 3* 5* 7*11*13*17*19*23*29*31*37*41*43*47*53 ... (End)
f:=n->mul(ithprime(k+1)^(binomial(n,k) mod 2),k=0..n); [seq(f(n),n=0..40)];
a[n_] := Product[Prime[k+1]^Mod[Binomial[n, k], 2], {k, 0, n}]; Table[a[n], {n, 0, 28}] (* Jean-François Alcover, Oct 01 2018, from Maple *)
a(n) = prod (k=0, n, if (binomial(n,k)%2, prime(k+1), 1)) \\ Rémy Sigrist, Jun 09 2019
from operator import mul from functools import reduce from sympy import prime def A123098(n): return reduce(mul,(1 if ~(n-1) & k else prime(k+1) for k in range(n))) # Chai Wah Wu, Feb 08 2016
(define (A123098 n) (A019565 (A001317 n))) ;; Antti Karttunen, Sep 18 2016
a(3) = 55 = (1, 3, 3, 1) dot (1, 2, 6, 30) = (1 + 6 + 18 + 30), where A002110 = (1, 2, 6, 30, 210, 2310, ...).
b:= proc(n) option remember; `if`(n=0, 1, ithprime(n)*b(n-1)) end: a:= n-> add(binomial(n, k)*b(k), k=0..n): seq(a(n), n=0..20); # Alois P. Heinz, Sep 20 2016
b[n_] := b[n] = If[n==0, 1, Prime[n]*b[n-1]]; a[n_] := Sum[Binomial[n, k]*b[k], {k, 0, n}]; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Feb 22 2017, translated from Maple *)
For n = 7 = prime(4), the corresponding polynomial is x^3, which factorizes as (x)(x)(x), thus a(7) = 3. For n = 14 = prime(4) * prime(1), the corresponding polynomial is x^3 + 1, which factorizes as (x + 1)(x^2 - x + 1), thus a(14) = 2. For n = 90 = prime(3) * prime(2)^2 * prime(1), the corresponding polynomial is x^2 + 2x + 1, which factorizes as (x + 1)^2, thus a(90) = 2. pfps(660) = pfps(2^2*3*5*11) = pfps(2^2) + pfps(3) + pfps(5) + pfps(11) = 2 + x + x^2 + x^4 which is irreducible, so a(660) = 1. For n = 30030 = Product_{i=1..6} prime(i), the corresponding polynomial is x^5 + x^4 + x^3 + x^2 + x + 1, which factorizes as (x+1)(x^2 - x + 1)(x^2 + x + 1), thus a(30030) = 3.
allocatemem(2^29); A064989(n) = {my(f); f = factor(n); if((n>1 && f[1,1]==2), f[1,2] = 0); for (i=1, #f~, f[i,1] = precprime(f[i,1]-1)); factorback(f)}; pfps(n) = if(1==n, 0, if(!(n%2), 1 + pfps(n/2), 'x*pfps(A064989(n)))); A277322 = n -> if(!bitand(n,(n-1)), 0, vecsum(factor(pfps(n))[,2])); for(n=1, 121121, write("b277322.txt", n, " ", A277322(n)));
pfps(n)=my(f=factor(n)); sum(i=1,#f~, f[i,2] * 'x^(primepi(f[i,1])-1)) A277322(n) = if(1==n, 0, vecsum(factor(pfps(n))[, 2])); \\ Charles R Greathouse IV, test for one added by Antti Karttunen, Oct 09 2016
A003961(n) = { my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); }; A242378(k,n) = { while(k>0,n = A003961(n); k = k-1); n; }; A285101(n) = { if(0==n,2,A285101(n-1)*A242378(n,A285101(n-1))); };
from sympy import factorint, prime, primepi from operator import mul from functools import reduce def a003961(n): f=factorint(n) return 1 if n==1 else reduce(mul, [prime(primepi(i) + 1)**f[i] for i in f]) def a242378(k, n): while k>0: n=a003961(n) k-=1 return n l=[2] for n in range(1, 7): x=l[n - 1] l.append(x*a242378(n, x)) print(l) # Indranil Ghosh, Jun 27 2017
(definec (A285101 n) (if (zero? n) 2 (* (A285101 (- n 1)) (A242378bi n (A285101 (- n 1)))))) ;; For A242378bi see A242378.
Terms are obtained by exponentiating the odd primes in range [3 .. prime(2+n)] with the binomial coefficients obtained from row n of Pascal's triangle (A007318) and then multiplying the factors together: 3^1 3^1 * 5^1 3^1 * 5^2 * 7^1 3^1 * 5^3 * 7^3 * 11^1 3^1 * 5^4 * 7^6 * 11^4 * 13^1 etc.
A276080 := proc (n) add((n-2*k)*factorial(n-k-1)/factorial(k), k = 0..floor((1/2)*n-1/2)) end proc: seq(A276080(n), n = 0..25); # Peter Bala, Dec 24 2017
Map[If[# == 1, 0, Total[FactorInteger[#] /. {p_, e_} /; p > 1 :> e PrimePi[p]!]] &, Nest[Append[#, (Times @@ Map[#1^#2 & @@ # &, FactorInteger[#] /. {p_, e_} /; e > 0 :> {Prime[PrimePi@ p + 1], e}] - Boole[# == 1] &@ #[[-1]]) #[[-2]]] &, {1, 2}, 24]] (* Michael De Vlieger, Dec 24 2017 *)
from sympy import factorint, factorial as f, prime, primepi from operator import mul from functools import reduce def a003961(n): F=factorint(n) return 1 if n==1 else reduce(mul, [prime(primepi(i) + 1)**F[i] for i in F]) def a276075(n): F=factorint(n) return 0 if n==1 else sum([F[i]*f(primepi(i)) for i in F]) l=[1, 2] L=[0, 1] for n in range(2, 11): l.append(a003961(l[n - 1])*l[n - 2]) L.append(a276075(l[n])) print(L) # Indranil Ghosh, Jun 21 2017
(define (A276080 n) (A276075 (A206296 n))) ;; A more practical standalone program, that uses memoization-macro definec: (define (A276080 n) (sum_factorials_times_elements_in (A206296as_index_lists n))) (definec (A206296as_index_lists n) (cond ((zero? n) (list)) ((= 1 n) (list 1)) (else (map + (cons 0 (A206296as_index_lists (- n 1))) (append (A206296as_index_lists (- n 2)) (list 0 0)))))) (define (sum_factorials_times_elements_in nums) (let loop ((s 0) (nums nums) (i 2) (f 1)) (cond ((null? nums) s) (else (loop (+ s (* (car nums) f)) (cdr nums) (+ 1 i) (* i f))))))
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