A159578 -id:A159578 - cp's OEIS frontend

A140762 Erroneous version of A159578.

30, 9699690, 304250263527210, 267064515689275851355624017992790, 5766152219975951659023630035336134306565384015606066319856068810, 962947420735983927056946215901134429196419130606213075415963491270

Offset: 1

dead

A051838 Numbers k such that sum of first k primes divides product of first k primes.

Original entry on oeis.org

1, 3, 8, 13, 23, 38, 39, 41, 43, 48, 50, 53, 56, 57, 58, 66, 68, 70, 73, 77, 84, 90, 94, 98, 126, 128, 134, 140, 143, 145, 149, 151, 153, 157, 160, 164, 167, 168, 172, 174, 176, 182, 191, 194, 196, 200, 210, 212, 215, 217, 218, 219, 222, 225, 228, 229

Offset: 1

G. L. Honaker, Jr., Dec 12 1999

nonn

			Sum of first 8 primes is 77 and product of first 8 primes is 9699690. 77 divides 9699690 therefore a(3)=8.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from T. D. Noe)

Cf. A007504, A002110, A159639, A196415.

A116536 gives the quotients, A140763 the divisors and A159578 the dividends.

P:=Filtered([1..2000],IsPrime);;
Filtered([1..Length(P)],n->Product([1..n],i->P[i]) mod Sum([1..n],i->P[i])=0); # Muniru A Asiru, Dec 20 2018

import Data.List (elemIndices)
a051838 n = a051838_list !! (n-1)
a051838_list =
   map (+ 1) $ elemIndices 0 $ zipWith mod a002110_list a007504_list
-- Reinhard Zumkeller, Oct 03 2011

p = Prime@ Range@ 250; Flatten@ Position[ Mod[ First@#, Last@#] & /@ Partition[ Riffle[ Rest[ FoldList[ Times, 1, p]], Accumulate@ p], 2], 0] (* Harvey P. Dale, Dec 19 2010 *)

for(n=1,100,P=prod(i=1,n,prime(i));S=sum(i=1,n,prime(i));if(!(P%S),print1(n,", "))) \\ Derek Orr, Jul 19 2015

isok(n) = my(p = primes(n)); (vecprod(p) % vecsum(p)) == 0; \\ Michel Marcus, Dec 20 2018

From Reinhard Zumkeller, Oct 03 2011: (Start)
A002110(a(n)) mod A007504(a(n)) = 0.
A116536(n) = A002110(a(n)) / A007504(a(n)). (End)

A116536 Numbers that can be expressed as the ratio of the product and the sum of consecutive prime numbers starting from 2.

Original entry on oeis.org

1, 3, 125970, 1278362451795, 305565807424800745258151050335, 2099072522743338791053378243660769678400212601239922213271230, 330455532167461882998265688366895823334392289157931734871641555

Offset: 1

Paolo P. Lava & Giorgio Balzarotti, Mar 27 2006

Let prime(i) denote the i-th prime (A000040). Let F(m) = (Product_{i=1..m} prime(i)) / (Sum_{i=1..m} prime(i)). Sequence gives integer values of F(m) and A051838 gives corresponding values of m. - N. J. A. Sloane, Oct 01 2011

			a(1) = 1 because 2/2 = 1.
a(2) = 3 because (2*3*5)/(2+3+5) = 30/10 = 3.
a(3) = 125970 because (2*3*5*7*11*13*17*19)/(2+3+5+7+11+13+17+19) = 9699690/77 = 125790.

G. Balzarotti and P. P. Lava, Le sequenze di numeri interi, Hoepli, 2008, p. 158.

Amiram Eldar, Table of n, a(n) for n = 1..81 (terms 1..42 from Vincenzo Librandi)

Cf. A001414, A108552, A067111, A051838, A140763, A141092, A159578.

Subsequence of A325307.

import Data.Maybe (catMaybes)
a116536 n = a116536_list !! (n-1)
a116536_list = catMaybes $ zipWith div' a002110_list a007504_list where
   div' x y | m == 0    = Just x'
            | otherwise = Nothing where (x',m) = divMod x y
-- Reinhard Zumkeller, Oct 03 2011

[p/s: n in [1..40] | IsDivisibleBy(p,s) where p is &*[NthPrime(i): i in [1..n]] where s is &+[NthPrime(i): i in [1..n]]];  // Bruno Berselli, Sep 30 2011

P:=proc(n) local i,j, pp,sp; pp:=1; sp:=0; for i from 1 by 1 to n do pp:=pp*ithprime(i); sp:=sp+ithprime(i); j:=pp/sp; if j=trunc(j) then print(j); fi; od; end: P(100);

seq = {}; sum = 0; prod = 1; p = 1; Do[p = NextPrime[p]; prod *= p; sum += p; If[Divisible[prod, sum], AppendTo[seq, prod/sum]], {50}]; seq (* Amiram Eldar, Nov 02 2020 *)

a(n) = A002110(A051838(n)) / A007504(A051838(n)). - Reinhard Zumkeller, Oct 03 2011

a(n) = A159578(n)/A001414(A159578(n)). - Amiram Eldar, Nov 02 2020

A141092 Product of first k composite numbers divided by their sum, when the result is an integer.

Original entry on oeis.org

1, 64, 46080, 111974400, 662171811840, 310393036800000, 7230916185292800, 108238138194410864640000, 23835710455777670400935290994688000000000, 1104077556971139123493322971152384000000000

Offset: 1

Enoch Haga, Jun 01 2008

Find the products and sums of first k composites, k = 1, 2, 3, .... When the products divided by the sums produce integral quotients, add terms to sequence.

			a(3)=46080 because 4*6*8*9*10*12*14=2903040 and 4+6+8+9+10+12+14=63; 2903040/63=46080, which is an integer, so 46080 is a term.

Reinhard Zumkeller, Table of n, a(n) for n = 1..90

Cf. A196415, A141089, A141090, A141091.
Compare with A140761, A159578, A140763, A116536.
Cf. A116536.

import Data.Maybe (catMaybes)
a141092 n = a141092_list !! (n-1)
a141092_list = catMaybes $ zipWith div' a036691_list a053767_list where
   div' x y | m == 0    = Just x'
            | otherwise = Nothing where (x',m) = divMod x y
-- Reinhard Zumkeller, Oct 03 2011

With[{cnos=Select[Range[50],CompositeQ]},Select[Table[Fold[ Times,1,Take[ cnos,n]]/ Total[Take[cnos,n]],{n,Length[cnos]}],IntegerQ]] (* Harvey P. Dale, Jan 14 2015 *)

s=0;p=1;forcomposite(n=4,100,p*=n;s+=n;if(p%s==0,print1(p/s", "))) \\ Charles R Greathouse IV, Apr 04 2013

a(n) = A036691(A196415(n)) / A053767(A196415(n)). [Reinhard Zumkeller, Oct 03 2011]

Checked by N. J. A. Sloane, Oct 02 2011.

A140763 A051838 gives numbers m such that the sum of first m primes divides the product of the first m primes. This sequence gives corresponding values of the sum of first m primes.

Original entry on oeis.org

2, 10, 77, 238, 874, 2747, 2914, 3266, 3638, 4661, 5117, 5830, 6601, 6870, 7141, 9523, 10191, 10887, 11966, 13490, 16401, 19113, 21037, 23069, 40313, 41741, 46191, 50887, 53342, 54998, 58406, 60146, 61910, 65534, 68341, 72179, 75130, 76127, 80189, 82253

Offset: 1

Enoch Haga, May 28 2008

nonn

Sums (divisors) associated with A140761.

			a(2)=10 because when 30 is divided by 10, the quotient is 3 and integral.

Amiram Eldar, Table of n, a(n) for n = 1..1000

Cf. A051838, A116536, A140761, A159578, A116536.

Module[{nn=200,s,p},s=Accumulate[Prime[Range[nn]]];p=FoldList[ Times,Prime[ Range[nn]]];Select[Thread[{p,s}],Divisible[#[[1]],#[[2]]]&]][[All,2]] (* Harvey P. Dale, Jun 07 2022 *)

a(n)=A116536(n)/A159578(n) = A007504(A051838(n)). - R. J. Mathar, Jun 09 2008

Sum_{i=1..A051838(n)} prime(i).

Corrected and edited by N. J. A. Sloane, Oct 01 2011

A140761 Primes p(j) = A000040(j), j>=1, such that p(1)p(2)...*p(j) is an integral multiple of p(1)+p(2)+...+p(j).

Original entry on oeis.org

2, 5, 19, 41, 83, 163, 167, 179, 191, 223, 229, 241, 263, 269, 271, 317, 337, 349, 367, 389, 433, 463, 491, 521, 701, 719, 757, 809, 823, 829, 859, 877, 883, 919, 941, 971, 991, 997, 1021, 1033, 1049, 1091, 1153, 1181, 1193, 1223, 1291, 1301, 1319, 1327, 1361

Offset: 1

Enoch Haga, May 28 2008

			a(2) = 5 because it is the last consecutive prime in the run 2*3*5 = 30 and 2+3+5 = 10; since 30/10 = 3, it is the first integral quotient.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A002110, A007504, A051838, A116536, A140763, A159578.

seq = {}; sum = 0; prod = 1; p = 1; Do[p = NextPrime[p]; prod *= p; sum += p; If[Divisible[prod, sum], AppendTo[seq, p]], {200}]; seq (* Amiram Eldar, Nov 02 2020 *)

Find integral quotients of products of consecutive primes divided by their sum.

a(n) = A000040(A051838(n)). - R. J. Mathar, Jun 09 2008

Edited by R. J. Mathar, Jun 09 2008

a(1) added by Amiram Eldar, Nov 02 2020

A141090 Integral quotients of products of first k consecutive composites divided by their sums: products (dividends).

Original entry on oeis.org

4, 1728, 2903040, 12541132800, 115880067072000, 69528040243200000, 1807729046323200000, 43295255277764345856000000, 20188846756043686829592191472500736000000000, 989253491046140654650017382152536064000000000

Offset: 1

Enoch Haga, Jun 01 2008

Based on A141092.
Compare with A140761 A159578 A140763 A116536.
Take the first k composite numbers. If their product divided by their sum results in an integer, their product is a term of the sequence. - Harvey P. Dale, Apr 29 2018

			a(3) = 2903040 because 4*6*8*9*10*12*14 = 2903040 and 4+6+8+9+10+12+14 = 63; 2903040/63 = 46080, integral -- 2903040 is added to the sequence.

Amiram Eldar, Table of n, a(n) for n = 1..93

Cf. A196415, A141089, A141091, A141092.

With[{c=Select[Range[100],CompositeQ]},Table[If[IntegerQ[ Times@@Take[ c,n]/Total[ Take[ c,n]]], Times@@ Take[ c,n],0],{n,Length[c]}]]/.(0-> Nothing) (* Harvey P. Dale, Apr 29 2018 *)

Find the products and sums of first k consecutive composites. When the product divided by the sum produces an integral quotient, add product to sequence.

Checked by N. J. A. Sloane, Oct 02 2011

Edited by N. J. A. Sloane, Apr 29 2018

A141091 Integral quotients of products of consecutive composites divided by their sums: sums (divisors).

Original entry on oeis.org

4, 27, 63, 112, 175, 224, 250, 400, 847, 896, 2368, 2448, 2695, 3596, 4300, 4624, 4961, 5076, 5546, 6032, 6156, 6664, 8750, 9048, 9200, 9976, 10295, 11620, 12312, 13572, 14697, 15872, 16275, 18139, 18352, 23572, 24304, 25544, 26814, 27072, 29986

Offset: 1

Enoch Haga, Jun 01 2008

			a(3) = 63 because 4*6*8*9*10*12*14 = 2903040 and 4+6+8+9+10+12+14 = 63; 2903040/63 = 46080, integral -- 63 is added to the sequence.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A196415, A141089, A141090, A141092.

Compare with A140761, A159578, A140763, A116536.

comp = Select[Range[500], CompositeQ]; sum = Accumulate[comp]; sum[[Position[ Rest@FoldList[Times, 1, comp]/sum, ?IntegerQ] // Flatten]] (* _Amiram Eldar, Jan 12 2020 *)

Find the products and sums of consecutive composites. When the products divided by the sums produce integral quotients, add terms to sequence.

a(37) corrected by Amiram Eldar, Jan 12 2020

A141089 Integral quotients of products of consecutive composites divided by their sums: Last consecutive composite.

Original entry on oeis.org

4, 9, 14, 18, 22, 25, 26, 33, 48, 49, 78, 80, 84, 95, 105, 110, 114, 115, 119, 123, 124, 129, 147, 150, 152, 158, 160, 170, 175, 184, 190, 200, 202, 212, 213, 242, 245, 250, 256, 258, 272, 284, 287, 288, 291, 306, 309, 314, 319, 327, 332, 333, 336, 342, 343

Offset: 1

Enoch Haga, Jun 01 2008

			a(3) = 14 because 4*6*8*9*10*12*14 = 2903040 and 4+6+8+9+10+12+14 = 63; 2903040/63 = 46080, integral -- 14 is added to the sequence.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A141090, A141091, A141092.

Compare with A140761, A159578, A140763, A116536.

comp = Select[Range[500], CompositeQ]; comp[[Position[Rest @ FoldList[Times, 1, comp]/Accumulate[comp], ?IntegerQ] // Flatten]] (* _Amiram Eldar, Jan 12 2020 *)

Find the products and sums of consecutive composites. When the products divided by the sums produce integral quotients, add terms to sequence.

A159580 Integral quotients occur consecutively this many times (in sequence associated with A051838 and A116536).

Original entry on oeis.org

2, 3, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 3, 2, 2, 2, 3, 2, 2, 3, 3, 2, 3, 2, 4, 4, 2, 2, 2, 2, 2, 2, 3, 2, 2, 3, 2, 2, 3, 3, 2, 2, 2, 2, 2, 3, 3, 2, 3, 2, 2, 3, 2, 3, 3, 2, 2, 2, 3, 2, 2, 3, 2, 2, 2, 5, 2, 2, 2, 2, 2, 2, 2, 4, 2

Offset: 1

Enoch Haga, Apr 16 2009

			a(1)=2 because the 1st subsequence (38,39) of consecutive integers has run length 2.
a(2)=3 because the 2nd subsequence is (56,57,58) which has run length 3.
a(3)=2 because the 3rd subsequence is (167,168) which has run length 2.

Cf. A051838, A116536, A140763, A159578, A159581.

Recomputed based on b-file of A051838. - R. J. Mathar, Aug 28 2018

cp's OEIS Frontend

A140762 Erroneous version of A159578.

Views

Author

Keywords

A051838 Numbers k such that sum of first k primes divides product of first k primes.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

GAP

Haskell

Mathematica

PARI

PARI

Formula

A116536 Numbers that can be expressed as the ratio of the product and the sum of consecutive prime numbers starting from 2.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Haskell

Magma

Maple

Mathematica

Formula

A141092 Product of first k composite numbers divided by their sum, when the result is an integer.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Mathematica

PARI

Formula

Extensions

A140763 A051838 gives numbers m such that the sum of first m primes divides the product of the first m primes. This sequence gives corresponding values of the sum of first m primes.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A140761 Primes p(j) = A000040(j), j>=1, such that p(1)*p(2)*...*p(j) is an integral multiple of p(1)+p(2)+...+p(j).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A141090 Integral quotients of products of first k consecutive composites divided by their sums: products (dividends).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

A140761 Primes p(j) = A000040(j), j>=1, such that p(1)p(2)...*p(j) is an integral multiple of p(1)+p(2)+...+p(j).