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.

Previous Showing 61-66 of 66 results.

A112438 Next term is the sum of the last 10 digits in the sequence, beginning with a(10) = 7.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 7, 7, 14, 19, 29, 40, 44, 38, 44, 42, 37, 43, 42, 37, 39, 45, 44, 45, 48, 50, 43, 41, 38, 40, 32, 32, 30, 28, 27, 32, 32, 32, 34, 31, 26, 29, 35, 38, 42, 44, 44, 41, 38, 38, 43, 42, 40, 39, 40, 33, 32, 31, 31, 23, 24, 24, 25, 28, 34, 36, 39, 45, 47, 48
Offset: 1

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Author

Alexandre Wajnberg, Dec 11 2005

Keywords

Comments

Variation on Angelini's A112395. The sequence cycles at a(18)=44 and the loop has 312 terms. Computed by Gilles Sadowski.

Examples

			a(18)=44 because 1+9 + 2+9 + 4+0 + 4+4 + 3+8 = 44

Crossrefs

Cf. A112395, A004207, A065075, A001370.

A112439 Next term is the sum of the last 10 digits in the sequence, beginning with a(10) = 8.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 8, 8, 16, 23, 28, 38, 49, 46, 49, 57, 59, 62, 57, 59, 60, 54, 49, 54, 51, 43, 44, 43, 37, 38, 43, 43, 42, 41, 36, 34, 34, 34, 35, 38, 40, 37, 40, 37, 39, 40, 40, 34, 37, 37, 35, 39, 47, 51, 47, 48, 52, 47, 47, 52, 48, 48, 53, 50, 44, 45, 42, 36, 37, 42
Offset: 1

Views

Author

Alexandre Wajnberg, Dec 11 2005

Keywords

Comments

Variation on Angelini's A112395. The sequence cycles at a(32)=37 and the loop has 312 terms. Computed by Gilles Sadowski.

Examples

			a(32)=37 because 5+4 + 5+1 + 4+3 + 4+4 + 4+3 = 37

Crossrefs

Cf. A112395, A004207, A065075, A001370.

A112440 Next term is the sum of the last 10 digits in the sequence, beginning with a(10) = 9.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 9, 9, 18, 27, 36, 45, 54, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45
Offset: 1

Views

Author

Alexandre Wajnberg, Dec 11 2005

Keywords

Comments

Variation on Angelini's A112395. The sequence cycles at a(17)=45 and the loop has one term. Computed by Gilles Sadowski.

Examples

			a(17)=45 because 1+8 + 2+7 + 3+6 + 4+5 + 5+4 = 45

Crossrefs

Cf. A112395, A004207, A065075, A001370.

A122063 a(1) = 17, a(n) = sum of digits of all previous terms.

Original entry on oeis.org

17, 8, 16, 23, 28, 38, 49, 62, 70, 77, 91, 101, 103, 107, 115, 122, 127, 137, 148, 161, 169, 185, 199, 218, 229, 242, 250, 257, 271, 281, 292, 305, 313, 320, 325, 335, 346, 359, 376, 392, 406, 416, 427, 440, 448, 464, 478, 497, 517, 530, 538, 554, 568, 587
Offset: 1

Views

Author

Herman Jamke (hermanjamke(AT)fastmail.fm), Sep 14 2006

Keywords

Comments

From a quiz.

References

  • A. Wareham, Test Your Brain Power, Ward Lock Ltd (1995).

Crossrefs

Cf. A004207.

Programs

  • Mathematica
    s={17};Do[AppendTo[s,Total[Total/@IntegerDigits/@s]],{n,53}];s (* James C. McMahon, Oct 25 2024 *)

A292566 Primes that can be reached with their prime-index, if you start with the prime-index and use iterations of A062028.

Original entry on oeis.org

2, 11, 107, 191, 587, 719, 1061, 1171, 1181, 1259, 1327, 1487, 1597, 1619, 1933, 1949, 2011, 2141, 2269, 2477, 2803, 2999, 3041, 3049, 3079, 3169, 3229, 3259, 3617, 3733, 4493, 4799, 5009, 5023, 5171, 5261, 5581, 5657, 6131, 6211, 6301, 6311, 6421, 6451, 6529
Offset: 1

Views

Author

Peter Weiss, Sep 19 2017

Keywords

Comments

If p is in the sequence, its index A000720(p) is not divisible by 3. - Robert Israel, Sep 19 2017

Examples

			The prime-index of 11 is 5: 5+5=10, 10+1+0=11 -> after two iterations you reach 11, so 11 is in the sequence.

Crossrefs

Cf. A000720, A062028, A007618, A004207.

Programs

  • Maple
    f:= proc(n) local t, p;
  p:= ithprime(n);
  t:= n;
  do
    t:= t + convert(convert(t,base,10),`+`);
    if t > p then return NULL
    elif t = p then return p
    fi
  od;
end proc:
map(f, [$1..1000]); # Robert Israel, Sep 19 2017
  • Mathematica
    ok[p_] := Block[{n = PrimePi@ p}, While [n < p, n += Total@ IntegerDigits@ n]; n == p]; Select[Prime@ Range@ 600, ok] (* Giovanni Resta, Sep 19 2017 *)
  • PARI
    is(n) = my(x=primepi(n)); while(1, x=x+sumdigits(x); if(x==n, return(1), if(x > n, return(0))))
forprime(p=1, 7000, if(is(p), print1(p, ", "))) \\ Felix Fröhlich, Sep 19 2017

Extensions

More terms from Felix Fröhlich, Sep 19 2017

A292568 a(n) = a(n-1) + sum of base-1000 digits of a(n-1), a(0)=1.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 1049, 1099, 1199, 1399, 1799, 2599, 3200, 3403, 3809, 4621, 5246, 5497, 5999, 7003, 7013, 7033, 7073, 7153, 7313, 7633, 8273, 8554, 9116, 9241, 9491, 9991, 10991, 11992, 12995, 14002, 14018, 14050, 14114, 14242, 14498
Offset: 0

Views

Author

Peter Weiss, Sep 19 2017

Keywords

Comments

In Germany you just write Q3 for the base-1000 digit sum (see book: "Taschenbuch der Mathematik" by Bronstein, Semendjajew, Musiol, Mühlig, p. 332) and you need it for the so-called "Teilbarkeitskriterium" for the number 37. If you add Q3 to a number you can also find this rule for the number 37.
Sum of base-1000 digits of m can also be described as "break the digit-string of m into triples starting at the right, and add these 3-digit numbers". For example, 1234567 -> 567 + 234 + (00)1 = 802.
None of the numbers of this sequence is divisible by 3 or 37.
The general form of this sequence is n + sum of base-(10^m) digits of n.
m=1: 1, 2, 4, 8, 16, 23, 28, 38, 49, 62, 70, 77, 91, 101, 103, ... (Cf. A004207.)
m=2: 1, 2, 4, 8, 16, 32, 64, 128, 157, 215, 232, 266, 334, 371, ... (Cf. A286660.)
m=3: 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 1049, 1099, ... (this sequence)

Examples

			a(16) = 2599 = 2 * 1000^1 + 599 * 1000^0. The sum of digits of a(17 - 1) = 2599 in base 1000 is therefore 2 + 599 = 601. a(17) = a(16) + the sum of digits of a(60) in base 1000 is therefore 2599 + 601 = 3200.

Crossrefs

Cf. A004207, A286660.

Programs

  • Mathematica
    NestList[Total[IntegerDigits[#,1000]]+#&,1,50] (* Harvey P. Dale, Dec 12 2018 *)
  • PARI
    a(n) = if (n==0, 1, prev = a(n-1); prev + sumdigits(prev, 1000)); \\ Michel Marcus, Sep 20 2017
Previous Showing 61-66 of 66 results.

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