A057003 -id:A057003 - cp's OEIS frontend

A336502 Partial sums of A057003.

1, 7, 127, 5167, 365527, 39435607, 6006997207, 1226103906007, 322796982334807, 106460296033918807, 42980408446129381207, 20846482682939051365207, 11959807608801430284133207, 8010447502346968140207973207, 6193994326661240674349352805207, 5476021766725276671842502543205207

Offset: 1

Seiichi Manyama, Jul 23 2020

Inspired by doubly triangular numbers (A002817).

			a(2) = 1 + 2*3 = 7.
a(3) = 1 + 2*3 + 4*5*6 = 127.
a(4) = 1 + 2*3 + 4*5*6 + 7*8*9*10 = 5167.

Seiichi Manyama, Table of n, a(n) for n = 1..226

Cf. A000217, A002817, A007489, A057003, A227364, A336513.

Accumulate @ Table[(n * (n + 1)/2)!/((n - 1) * n /2)!, {n, 1, 16}] (* Amiram Eldar, Jul 23 2020 *)

{a(n) = sum(i=1, n, prod(j=(i-1)*i/2+1, i*(i+1)/2, j))}

a(n) = Sum_{i=1..n} Product_{j=T(i-1)+1..T(i)} j where T(n) is n-th triangular number.
a(n) = A227364(T(n)) where T(n) is n-th triangular number.
a(n) ~ n^(2*n) / 2^n. - Vaclav Kotesovec, Nov 20 2021

A117384 Positive integers, each occurring twice in the sequence, such that a(n) = a(k) when n+k = 4*a(n), starting with a(1)=1 and filling the next vacant position with the smallest unused number.

Original entry on oeis.org

1, 2, 1, 3, 4, 2, 5, 3, 6, 7, 4, 8, 5, 9, 6, 10, 11, 7, 12, 8, 13, 9, 14, 10, 15, 16, 11, 17, 12, 18, 13, 19, 14, 20, 15, 21, 22, 16, 23, 17, 24, 18, 25, 19, 26, 20, 27, 21, 28, 29, 22, 30, 23, 31, 24, 32, 25, 33, 26, 34, 27, 35, 28, 36, 37, 29, 38, 30, 39, 31, 40, 32, 41, 33, 42

Offset: 1

Paul D. Hanna, Mar 11 2006

Positions where n occurs are A001614(n) and 4*n-A001614(n), where A001614 is the Connell sequence: 1 odd, 2 even, 3 odd, ...
From Paolo Xausa, Aug 27 2021: (Start)
Terms can be arranged in an irregular triangle T(r,c) read by rows in which row r is a permutation P of the integers in the interval [s, s+rlen-1], where s = 1+(r-1)*(r-2)/2, rlen = 2*r-1 and r >= 1 (see example).
P is the alternating (first term > second term < third term > fourth term ...) permutation m -> 1, 1 -> 2, m+1 -> 3, 2 -> 4, m+2 -> 5, 3 -> 6, ..., rlen -> rlen, where m = ceiling(rlen/2).
The triangle has the following properties.
Row lengths are the positive odd numbers.
First column is A000124.
Terms in column c (where c >= 1) are of the form k*(k+1)/2+ceiling(c/2), for integers k >= floor((c-1)/2), each even column being equal to the column preceding it.
Row records (the positive terms of A000217) are in the right border.
Indices of row records are the positive terms of A000290.
Each row r contains r terms that are duplicated in the next row.
In each row, the sum of terms which are not already listed in the sequence gives the positive terms of A006003.
Row sums give A063488.
For rows r >= 2, row product is A057003(r)*A057003(r-1). (End)

			9 first appears at position: A001614(9) = 14;
9 next appears at position: 4*9 - A001614(9) = 22.
From _Paolo Xausa_, Aug 27 2021: (Start)
Written as an irregular triangle T(r,c) the sequence begins:
  r\c  1   2   3   4   5   6   7   8   9  10  11  12  13
  1:   1;
  2:   2,  1,  3;
  3:   4,  2,  5,  3,  6;
  4:   7,  4,  8,  5,  9,  6, 10;
  5:  11,  7, 12,  8, 13,  9, 14, 10, 15;
  6:  16, 11, 17, 12, 18, 13, 19, 14, 20, 15, 21;
  7:  22, 16, 23, 17, 24, 18, 25, 19, 26, 20, 27, 21, 28;
  ...
The triangle can be arranged as shown below so that, in every row, each odd position term is equal to the term immediately below it.
               1
            2  1  3
         4  2  5  3  6
      7  4  8  5  9  6 10
  11  7 12  8 13  9 14 10 15
             ...
(End)

Cf. A117385 (a(5*a(n)-n)=a(n)), A117386 (a(6*a(n)-n)=a(n)).
Cf. A001614 (Connell sequence).
Cf. A000124, A000217, A006003, A057003, A063488, A344482.

nterms=64;a=ConstantArray[0,nterms];For[n=1;t=1,n<=nterms,n++,If[a[[n]]==0,a[[n]]=t;If[(d=4t-n)<=nterms,a[[d]]=a[[n]]];t++]]; a (* Paolo Xausa, Aug 27 2021 *)
(* Second program, triangle rows *)
nrows = 8;Table[rlen=2r-1;Permute[Range[s=1+(r-1)(r-2)/2,s+rlen-1],Join[Range[2,rlen,2],Range[1,rlen,2]]],{r,nrows}] (* Paolo Xausa, Aug 27 2021 *)

{a(n)=local(A=vector(n),m=1); for(k=1,n,if(A[k]==0,A[k]=m;if(4*m-k<=#A,A[4*m-k]=m);m+=1));A[n]}

T(r,c) = my(k = r-1-((c+1) % 2)); k*(k+1)/2+ceil(c/2);
tabf(nn) = {for (r=1, nn, for(c = 1, 2*r-1, print1(T(r,c), ", ");); print;);} \\ Michel Marcus, Sep 09 2021

a(4*a(n)-n) = a(n).
Lim_{n->infinity} a(n)/n = 1/2.
Lim_{n->infinity} (a(n+1)-a(n))/sqrt(n) = 1.
a( A001614(n) ) = n; a( 4n - A001614(n) ) = n.
T(r,c) = k*(k+1)/2+ceiling(c/2), where k = r-1-((c+1) mod 2), r >= 1 and c >= 1. - Paolo Xausa, Sep 09 2021

A062029 Group even numbers into (2), (4,6), (8,10,12), (14,16,18,20), ...; a(n) = product of n-th group.

Original entry on oeis.org

2, 24, 960, 80640, 11531520, 2500485120, 763847884800, 312344808652800, 164644289755545600, 108684799028822016000, 87805845811395506995200, 85211145316323008446464000, 97803969545162680178835456000, 131047222390590123375392194560000, 202702319752278628965061257854976000

Offset: 1

Amarnath Murthy, Jun 02 2001

nonn

			a(3) = 8*10*12 = 960.

Harry J. Smith, Table of n, a(n) for n=1..100

Cf. A057003, A062030, A062031, A062032.

Table[2^n*Gamma[(2+n+n^2)/2]/Gamma[(2-n+n^2)/2], {n,30}] (* G. C. Greubel, May 05 2022 *)
With[{nn=30},Times@@@TakeList[Range[2,(nn(nn+1))/2,2],Range[nn/2]]] (* Harvey P. Dale, May 09 2022 *)

a(n) = { 2^n*((n^2 + n)/2)!/((n^2 - n)/2)! } \\ Harry J. Smith, Jul 30 2009

[2^n*gamma((2+n+n^2)/2)/gamma((2-n+n^2)/2) for n in (1..30)] # G. C. Greubel, May 05 2022

a(n) = Product_{k=1..n} (n^2 - n + 2*k) = (n^2 + n)!!/(n^2 - n)!! .
a(n) = 2^n*Gamma((n^2 + n + 2)/2)/Gamma((n^2 - n + 2)/2).
a(n) = 2^n * A057003(n-1).

Formula and more terms from Vladeta Jovovic, Jun 05 2001

A093451 Number of distinct prime divisors of Product_{k=1+(n-1)n/2..n(n+1)/2} k (i.e., of 1, 23, 456, 78910, ...).

Original entry on oeis.org

0, 2, 3, 4, 6, 6, 7, 8, 10, 10, 11, 13, 13, 14, 16, 15, 18, 17, 20, 19, 22, 21, 22, 24, 24, 26, 26, 27, 30, 28, 30, 31, 32, 33, 33, 36, 35, 36, 38, 39, 39, 39, 43, 41, 43, 44, 44, 47, 45, 49, 48, 48, 52, 49, 53, 53, 54, 54, 55, 58, 55, 60, 59, 59, 62, 60, 65, 64, 64, 65, 66, 68

Offset: 1

Amarnath Murthy, Apr 03 2004

nonn

			a(7) = 7 as the prime divisors of the product 22*23*24*25*26*27*28 are 2,3,5,7,11,13 and 23.

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A001221, A057003.

with(numtheory): a:=n->nops(factorset(product(k,k=1+n*(n-1)/2..n*(n+1)/2))): seq(a(n),n=1..80); # Emeric Deutsch, Feb 05 2006

With[{nn=75},PrimeNu[#]&/@Times@@@TakeList[Range[(nn(nn+1))/2],Range[ nn]]] (* Harvey P. Dale, Sep 01 2021 *)

a(n) = { my(b=binomial(n,2)+1, bp1=binomial(n+1,2), res = primepi(n)); forprime(p = n + 1, bp1, bp = b%p; if(bp > bp1 % p || bp == 0, res++ ) ); res } \\ David A. Corneth, Sep 01 2021

a(n) = A001221(A057003(n)). - Michel Marcus, Jul 29 2017

Corrected and extended by Emeric Deutsch, Feb 05 2006

A093456 Product of composite numbers among next n numbers.

Original entry on oeis.org

1, 1, 24, 720, 2520, 120960, 259459200, 1357171200, 4929724800, 42608389824000, 11912739135897600, 59907396092544000, 20458385028297216000, 7926428532945162240000, 4693751193479184764928000, 328774885640356760904499200000, 12797917159224592605450240000

Offset: 1

Amarnath Murthy, Apr 03 2004

Conjecture: There are finitely many numbers such that a(n) is not == 0 (mod a(n-1)). (Also mentioned in A093455.)

Product of all composite numbers between n*(n-1)/2+1 and n*(n+1)/2 (including boundaries). - Stefan Steinerberger, Apr 02 2006

			Sequence begins:
   1:             a(1) = 1.
   2  3:          a(2) = 1.
   4  5  6:       a(3) = 4*6 = 24.
   7  8  9 10:    a(4) = 8*9*10 = 720.
  11 12 13 14 15: a(5) = 12*14*15 = 2520.
  ...

Harvey P. Dale, Table of n, a(n) for n = 1..244

Cf. A000217, A057003, A093455, A093457.

Table[a := Range[n*(n - 1)/2 + 1, n*(n + 1)/2]; b := Select[a, Not[PrimeQ[ # ]] &]; Product[b[[i]], {i, 1, Length[b]}], {n, 1, 20}] (* Stefan Steinerberger, Apr 02 2006 *)
Module[{nn=20},Times@@Select[#,CompositeQ]&/@TakeList[Range[(nn(nn+1))/2],Range[nn]]] (* Harvey P. Dale, Dec 30 2024 *)

a(n) = A057003(n)/A093457(n). - Michel Marcus, Jan 14 2025

More terms from Stefan Steinerberger, Apr 02 2006

A062079 Group the odd numbers as (1), (3,5), (7,9,11), (13,15,17,19), (21,23,25,27,29), ... then a(n) = LCM of the n-th group.

Original entry on oeis.org

1, 15, 693, 62985, 3151575, 706110405, 44166438855, 30637289555145, 3274769391079725, 312250034062131165, 593968671422526274875, 5531265959247033940935, 95840860214492177176316925

Offset: 1

Amarnath Murthy, Jun 15 2001

nonn

			a(3) = lcm(7,9,11) = 693.

Harry J. Smith, Table of n, a(n) for n = 1..100

Cf. A057003, A062032. - Franklin T. Adams-Watters, Jul 03 2009

Table[LCM[Gamma[2*Binomial[n+1, 2] + 1]*Gamma[Binomial[n, 2] + 1]/(2^n*Gamma[Binomial[n+1, 2] + 1]*Gamma[2*Binomial[n, 2] + 1])], {n,20}] (* G. C. Greubel, May 13 2022 *)

a(n) = local(r);r=1;forstep(k=n^2-n+1,n^2+n-1,2,r=lcm(r,k));r \\ Franklin T. Adams-Watters, Jul 03 2009

{ for (n=1, 100, a=b=n^2 - n + 1; for (k=1, n - 1, a=lcm(a, b + 2*k)); write("b062079.txt", n, " ", a) ) } \\ Harry J. Smith, Jul 31 2009

[lcm(gamma(2*binomial(n+1, 2) + 1)*gamma(binomial(n, 2) + 1)/(2^n*gamma(binomial(n+1, 2) + 1)*gamma(2*binomial(n, 2) + 1))) for n in (1..20)] # G. C. Greubel, May 13 2022

a(n) = lcm(Gamma(2*binomial(n+1, 2) + 1)*Gamma(binomial(n, 2) + 1)/(2^n*Gamma(binomial(n+1, 2) + 1)*Gamma(2*binomial(n, 2) + 1))). - G. C. Greubel, May 13 2022

Corrected and extended by Franklin T. Adams-Watters, Jul 03 2009

A071226 n-th power of the product of next n natural numbers.

Original entry on oeis.org

1, 36, 1728000, 645241282560000, 6076911214672415134617600000, 3556852067865008593425339325122707718144000000, 269512166306728282203388439200933086924875888324090265600000000000000

Offset: 1

Amarnath Murthy, May 17 2002

nonn

Harvey P. Dale, Table of n, a(n) for n = 1..20

Cf. A057003.

seq(product(n*(n-1)/2+i,i=1..n)^n,n=1..7);

Table[Pochhammer[1 + (n-1)*n/2, n]^n, {n, 1, 8}] (* Vaclav Kotesovec, Jul 13 2021 *)
With[{nn=10},(Times@@#)^Length[#]&/@TakeList[Range[(nn(nn+1))/2],Range[ nn]]] (* Harvey P. Dale, Aug 09 2021 *)

a(n) ~ exp(5/6) * n^(2*n^2) / 2^(n^2). - Vaclav Kotesovec, Jul 13 2021

More terms from Sascha Kurz, Jan 02 2003

A093908 Let f(k, n) be the product of n consecutive numbers beginning with k. Then a(n) is the least k > 1+n(n-1)/2 such that f(k, n) is a multiple of f(1+n(n-1)/2, n).

Original entry on oeis.org

2, 3, 8, 39, 52, 187, 204, 863, 773, 6621, 34038, 2404, 34440, 223097, 11976, 1106290, 1980047, 85119892, 15308072, 496820597, 2590416388, 1087065675, 4736428784, 1128909067, 242793786666, 2791304683100, 273924845940

Offset: 1

Amarnath Murthy, Apr 24 2004

nonn

f(k, n) = A008279(n+k-1, n). 1+n*(n-1)/2 = A000124(n-1). f(1+n*(n-1)/2, n) = A057003(n).

a(28) > 88*10^12.

			a(4) = 39 because 39*40*41*42 is divisible by 7*8*9*10. No
smaller set gives a product that is a multiple of 7*8*9*10.

Cf. A000124, A008279, A057003, A093909.

Edited and extended by David Wasserman, Apr 25 2007

A072529 Product of the next n multiples of n.

Original entry on oeis.org

1, 24, 3240, 1290240, 1126125000, 1822853652480, 4914543582748800, 20469829379869900800, 124583146969045247683200, 1061374990515840000000000000, 12232437772393129356474535526400

Offset: 1

Amarnath Murthy, Aug 01 2002

nonn

			a(4) = 10*4 * 9*4 * 8*4 * 7*4 = 1290240.

a(n) = A057003(n) * n^n.

a(n) = n^n * (n*(n+1)/2)! / (n*(n-1)/2)!; \\ Michel Marcus, Apr 16 2025

a(n) = n^n * (n*(n+1)/2)! / (n*(n-1)/2)!. - Lior Manor, Aug 19 2002

More terms from Lior Manor Aug 19 2002

A092934 a(n) = floor((product of next n even numbers) / (product of first n odd numbers)).

Original entry on oeis.org

2, 8, 64, 768, 12202, 240546, 5652480, 154090259, 4777917500, 165999652648, 6386199899437, 269455867248640, 12371082837260281, 613924958232961934, 32745240915899894988, 1868019304186661949347, 113491057175295931012181

Offset: 1

Amarnath Murthy, Mar 22 2004

Table[Floor[2^n(n (n + 1)/2)!/(((n - 1) n/2)!*(2n - 1)!!)], {n, 1, 20}] (* Stefan Steinerberger, Jan 28 2006 *)

a(n) = floor(2^n*A057003(n)/A001147(n)).

More terms from Stefan Steinerberger, Jan 28 2006

cp's OEIS Frontend

A336502 Partial sums of A057003.

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A117384 Positive integers, each occurring twice in the sequence, such that a(n) = a(k) when n+k = 4*a(n), starting with a(1)=1 and filling the next vacant position with the smallest unused number.

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A062029 Group even numbers into (2), (4,6), (8,10,12), (14,16,18,20), ...; a(n) = product of n-th group.

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A093451 Number of distinct prime divisors of Product_{k=1+(n-1)n/2..n(n+1)/2} k (i.e., of 1, 2*3, 4*5*6, 7*8*9*10, ...).

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A093456 Product of composite numbers among next n numbers.

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A062079 Group the odd numbers as (1), (3,5), (7,9,11), (13,15,17,19), (21,23,25,27,29), ... then a(n) = LCM of the n-th group.

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A071226 n-th power of the product of next n natural numbers.

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A093451 Number of distinct prime divisors of Product_{k=1+(n-1)n/2..n(n+1)/2} k (i.e., of 1, 23, 456, 78910, ...).

A093908 Let f(k, n) be the product of n consecutive numbers beginning with k. Then a(n) is the least k > 1+n(n-1)/2 such that f(k, n) is a multiple of f(1+n(n-1)/2, n).