cp's OEIS Frontend

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.

Showing 1-4 of 4 results.

A352346 Common terms between A061078 and A061077.

Original entry on oeis.org

26, 52, 148, 280, 320, 454, 1150, 1480, 8000, 41650, 80300, 165656, 166088, 614900, 2353700, 2859460, 28233200, 66130400, 68941640, 85717240, 107300320, 131507080, 155478800, 207666520, 1426680920, 1824596800, 2468014900, 2475648820, 5342351060, 5355218900, 5857281500, 8550475900, 36025361120
Offset: 1

Views

Author

Luca Onnis, Mar 12 2022

Keywords

Comments

Smarandache's conjecture: there are infinitely many terms.
This is a subsequence of A061076.

Examples

			26 is a term of this sequence, in fact:
26 = 1+3+5+7+9+1*1 (A061077(6)=26);
26 = 2+4+6+8+1*0+1*2+1*4 (A061078(7)=26).

References

  • A. Murthy, Smarandache friendly numbers and a few more sequences, Smarandache Notions Journal, Vol. 12, No. 1-2-3, Spring 2001. Page 267

Links

Crossrefs

Cf. A061076, A061077, A061078.

Programs

  • Mathematica
    Intersection[Accumulate[Times @@@ IntegerDigits[Range[2, 10000000, 2]]],
Accumulate[Times @@@ IntegerDigits[Range[1, 10000000, 2]]]]
  • Python
    from math import prod
from itertools import islice
def A352346_gen(): # generator of terms
    n1, m1, n2, m2 = 1, 1, 2, 2
    while True:
        if m1 == m2:
            yield m1
        k = 0
        while k == 0:
            n1 += 2
            m1 += (k := prod(int(d) for d in str(n1)))
        while m2 < m1:
            n2 += 2
            m2 += prod(int(d) for d in str(n2))
A352346_list = list(islice(A352346_gen(),20)) # Chai Wah Wu, Mar 21 2022

A350744 Numbers m such that A061078(m)/A061077(m) = 4/5.

Original entry on oeis.org

5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 51, 52, 53, 54, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100, 101, 102, 103, 104, 105, 110, 115, 120, 125, 130, 135, 140, 145, 150, 151, 152, 153, 154, 155, 160, 165, 170, 175, 180, 185, 190, 195, 200, 201, 202, 203, 204, 205, 210, 215, 220, 225
Offset: 1

Views

Author

Luca Onnis, Mar 20 2022

Keywords

Comments

All positive multiples of 5 are terms of the sequence.

Examples

			30 is a term, in fact A061078(30)=320, A061077(30)=400 and a(n) = 320/400 = is 4/5.
500, 501, 502, ..., 554, 555 are all terms. In fact 500=5*10^2 and for the formula above also 501, ..., 500+(5/9)*(10^2-1) = 555 are all terms of the sequence.

References

  • Amarnath Murthy, Smarandache friendly numbers and a few more sequences, Smarandache Notions Journal, Vol. 12, No. 1-2-3, Spring 2001.

Links

Crossrefs

Cf. A061076, A061077, A061078.

Programs

  • Mathematica
    Flatten[Position[(Accumulate[Times @@@ IntegerDigits[Range[2, 10000, 2]]]/
    Accumulate[Times @@@ IntegerDigits[Range[1, 9999, 2]]]), 4/5]]
  • PARI
    pd(n) = my(d = digits(n)); prod(i=1, #d, d[i]);
isok(k) = sum(i=1, k, pd(2*i))/sum(i=1, k, pd(2*i-1)) == 4/5; \\ Michel Marcus, Mar 21 2022

Formula

Let k be a positive integer not divisible by 5 and j >= 0; then 5*k*10^j, 5*k*10^j+1, ..., 5*k*10^j+(5/9)*(10^j-1) are all terms of the sequence.
Limit_{n->oo} A061078(n)/A061077(n) = 4/5.

A061078 Sum of the products of the digits of the first n positive even numbers.

Original entry on oeis.org

2, 6, 12, 20, 20, 22, 26, 32, 40, 40, 44, 52, 64, 80, 80, 86, 98, 116, 140, 140, 148, 164, 188, 220, 220, 230, 250, 280, 320, 320, 332, 356, 392, 440, 440, 454, 482, 524, 580, 580, 596, 628, 676, 740, 740, 758, 794, 848, 920, 920, 920, 920, 920, 920, 920, 922
Offset: 1

Views

Author

Amarnath Murthy, Apr 14 2001

Keywords

Comments

For n = (10^r)/2, a(n) is the sum of the r terms of the geometric progression with first term 20 and common ratio 45.

Examples

			a(5) = 2 + 4 + 6 + 8 + 1*0 = 20; (a(18)=116, not 114).
a(1199) = a(5^2*2^4*3 - 1) = ... = a(5^2*2^4*3 + 5) = a(1205). In fact, the number of "fives" is exactly equal to 1 = 2-1 (where 2 is the exponent of 5).

References

  • Amarnath Murthy, Smarandache friendly numbers and a few more sequences, Smarandache Notions Journal, Vol. 12, No. 1-2-3, Spring 2001.

Links

Crossrefs

Cf. A061076, A061077.

Programs

  • Mathematica
    Accumulate[Times@@@IntegerDigits[Range[2,120,2]]] (* Harvey P. Dale, Jun 18 2021 *)
  • PARI
    pd(n) = my(d = digits(n)); prod(i=1, #d, d[i]);
a(n) = sum(k=1, n, pd(2*k)); \\ Michel Marcus, Feb 01 2015
  • PARI
    a(n) = sum(k=1, n, vecprod(digits(2*k))); \\ Michel Marcus, Mar 13 2022
  • PARI
    a(n) = {t=digits(2*n); p=1; d=#t; for(i=1, #t, if(t[i]==0, d=i-1; break));
(5/11) * (45^(#t-1)-1) + (sum(i=1, #t-1, ((prod(j=1,#t-i-1,t[j])) * (t[#t-i]) * (t[#t-i]-1) * 2 * (5^(i))* (9^(i-1)))))+(prod(k=1,#t-1,t[k]))*((((t[#t])^2))/4+(t[#t])/2)} \\ Luca Onnis, Mar 17 2022
  • Python
    from math import prod
from itertools import accumulate
def p(n): return prod(map(int, str(n)))
def a(n): return sum(p(2*i) for i in range(1, n+1))
def aupton(nn): return list(accumulate([pd(2*k) for k in range(1, nn+1)]))
print(aupton(56)) # Michael S. Branicky, Mar 13 2022

Formula

From Luca Onnis, Mar 13 2022: (Start)
a(5*10^n-1) = a(5*10^n) = (5/11)*(45^(n+1)-1).
a(n) <= (5/11)*(45^(log((n+1)/5)+1)-1) for all n.
a(n) ~ (4/5)*A061077(n) as n -> infinity.
Conjecture: let a >= 1, b >= 0, where p is not a multiple of 2 nor 5. Then:
a(5^a*2^b*p-1) = a(5^a*2^b*p) = ... = a(5^a*2^b*p + 55...5) where the number of fives is equal to b if a > b, and is equal to a-1 if 1 <= a <= b. (End)

Extensions

Corrected and extended by Matthew Conroy, Apr 17 2001
Incorrect formula removed by Luca Onnis, Mar 13 2022
Name clarified by Chai Wah Wu, Mar 21 2022

A360365 a(n) = sum of the products of the digits of the first n positive multiples of 3.

Original entry on oeis.org

3, 9, 18, 20, 25, 33, 35, 43, 57, 57, 66, 84, 111, 119, 139, 171, 176, 196, 231, 231, 249, 285, 339, 353, 388, 444, 452, 484, 540, 540, 567, 621, 702, 702, 702, 702, 703, 707, 714, 714, 720, 732, 750, 756, 771, 795, 799, 815, 843, 843, 858, 888, 933, 945, 975, 1023, 1030, 1058, 1107
Offset: 1

Views

Author

Luca Onnis, Feb 09 2023

Keywords

Examples

			a(4) = 20 since the first 4 positive multiples of 3 are 3,6,9 and 12 and the sum of the product of their digits is 3 + 6 + 9 + 1*2 = 20.

Crossrefs

Cf. A061076, A061077, A061078.

Programs

  • Mathematica
    Accumulate[Times @@@ IntegerDigits[Range[3, 999 , 3]]]
  • PARI
    a(n)={sum(k=1, n, vecprod(digits(3*k)))} \\ Andrew Howroyd, Feb 09 2023
  • Python
    from math import prod
def A360365(n): return sum(prod(int(d) for d in str(m)) for m in range(3,3*n+1,3)) # Chai Wah Wu, Feb 28 2023

Formula

a((10^n-1)/3) = (1/836)*(15*(19*45^n-63) + 44*3^((3*n)/2)*(sqrt(3)*sin((Pi*n)/6) + 15*cos((Pi*n)/6))).
G.f. of the subsequence a((10^n-1)/3): 9*(2 - 32*x + 135*x^2)/((1 - x)*(1 - 45*x)*(1 - 9*x + 27*x^2)).
a((10^n-1)/3) = 55*a((10^(n-1)-1)/3) - 486*a((10^(n-2)-1)/3) + 1647*a((10^(n-3)-1)/3) - 1215*a((10^(n-4)-1)/3) for n > 4.
Showing 1-4 of 4 results.

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