A004207 -id:A004207 - cp's OEIS frontend

A248078 a(1) = 1; a(n+1) = a(n) + product of digits of a(n) + sum of digits of a(n).

1, 3, 9, 27, 50, 55, 90, 99, 198, 288, 434, 493, 617, 673, 815, 869, 1324, 1358, 1495, 1694, 1930, 1943, 2068, 2084, 2098, 2117, 2142, 2167, 2267, 2452, 2545, 2761, 2861, 2974, 3500, 3508, 3524, 3658, 4400, 4408, 4424, 4566, 5307, 5322, 5394, 5955, 7104, 7116

Offset: 1

Gil Broussard, Sep 30 2014

Unlike A063108, this sequence includes in its formula the digit 0 in the product of digits of a(n).

			Given a(5)=50, then a(6)=50+(5+0)+(5*0)=55.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A063108, A004207.

f:= proc(x) local L;
  L:= convert(x,base,10);
  x + convert(L,`+`)+convert(L,`*`)
end proc:
A[1]:= 1:
for n from 2 to 100 do A[n]:= f(A[n-1]) od:
seq(A[i],i=1..100); # Robert Israel, Jun 25 2019

NestList[#+Total[IntegerDigits[#]]+Times@@IntegerDigits[#]&,1,50] (* Harvey P. Dale, Sep 11 2019 *)

lista(nn) = {prev = 1; print1(prev, ", "); for (n=1, nn, d = digits(prev); prev += sumdigits(prev) + prod(k=1, #d, d[k]); print1(prev, ", "););} \\ Michel Marcus, Oct 01 2014

A272233 Number of steps before n1(i) = n2(i) when n1(i) = n1(i-1) + digsum(n2(i-1)), n2(i) = n2(i-1) + digsum(n1(i-1)) and n1(1) = 10^(n-1), n2(1) = 0.

Original entry on oeis.org

1, 12, 57, 22820, 754504

Offset: 0

Anthony Sand, Apr 23 2016

The sequence takes two different numbers, n1 and n2, and simultaneously adds the digit sum of n1 to n2 and the digit sum of n2 to n1. This process continues until n1 = n2. The two numbers are initialized with n1 = 10^(n-1) and n2 = 0.

a(5) > 10^12. - Lars Blomberg, Jul 19 2017

			10 > 0, 10 > 1, 11 > 2, 13 > 4, 17 > 8, 25 > 16, 32 > 23, 37 > 28, 47 > 38, 58 > 49, 71 > 62, 79 > 70, 86 = 86

Cf. A004207, A272235

{digsum(num) = d=digits(num,b); return(sum(i=1,#d,d[i]));} {doubledigsum() = b=10; nmx=5; for(n=1,nmx, n1=b^(a-1); n2=0; c=0; until(n1==n2, s1=digsum(n1); s2=digsum(n2); n1+=s2; n2+=s1; c++); print1(c,", "); ); }

n1(i) = n1(i-1) + digsum(n2(i-1)), n2(i) = n2(i-1) + digsum(n1(i-1))

A272235 In base 2, number of steps before n1(i) = n2(i) when n1(i) = n1(i-1) + digsum(n2(i-1)), n2(i) = n2(i-1) + digsum(n1(i-1)) and n1(1) = 2^(n-1), n2(1) = 0.

Original entry on oeis.org

1, 3, 5, 8, 1204, 1205, 1199, 1191, 19536395, 19536233, 19535912, 19535673, 19519159

Offset: 0

Anthony Sand, Apr 23 2016

The sequence takes two different binary numbers, n1 and n2, and simultaneously adds the digit sum of n1 to n2 and the digit sum of n2 to n1. This process continues until n1 = n2. The two numbers are initialized with n1 = 2^(n-1) and n2 = 0.

			In base 2: 1000 > 0, 1000 > 1, 1001 > 10, 1010 > 100, 1011 > 110, 11111 > 1100, 10001 > 10000, 10010 = 10010
In base 10: 8 > 0, 8 > 1, 9 > 2, 10 > 4, 11 > 6, 13 > 9, 15 > 12, 17 > 16, 18 = 18

Cf. A004207, A272233.

digsum(num) = d=digits(num,2); return(sum(i=1,#d,d[i]));
doubledigsum() = b=2; nnx=5; for(n=1,amx, n1=b^(n-1); n2=0; c=0; until(n1==n2, s1=digsum(n1); s2=digsum(n2); n1+=s2; n2+=s1; c++); print1(c,", "); );

n1(i) = n1(i-1) + digsum(n2(i-1),base=2), n2(i) = n2(i-1) + digsum(n1(i-1),base=2)

A289979 Define two sequences n1(i) and n2(i) by the recurrences n1(i) = n1(i-1) + digsum(n2(i-1)), n2(i) = n2(i-1) + digsum(n1(i-1)), with initial values n1(1) = n and n2(1) = 0. Then a(n) is the smallest m such that n1(i) = n2(i) = m for some i, or -1 if no such m exists.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 86, 86, 42, 86, 20, 42, 53, 86, 108, 20, 110, 222, 110, 31, 222, 310, 110, 288, 31, 97, 75, 154, 64, 75, 692, 154, 468, 64, 176, 75, 389, 367, 132, 187, 389, 648, 367, 209, 132, 211, 1772, 411, 446, 1715, 828, 1772, 7150, 411, 413

Offset: 1

Anthony Sand, Jul 17 2017

The function is like a chase that ends when n1(i) = n2(i). For example, when n = 14:
n1(1) = 14, n2(1) = 0
n1(2) = 14 = 14 + digsum(0), n2(2) = 5 = 0 + digsum(14)
n1(3) = 19 = 14 + digsum(5), n2(3) = 10 = 5 + digsum(14)
n1(4) = 20 = 19 + digsum(10), n2(4) = 20 = 10 + digsum(19)
Because n1 = n2 = 20, the chase ends and a(14) = 20. When n = 81, a(81) > 10^8 and the chase may never end. In other bases, some different number first produces a prolonged chase with no result. E.g., in base 9, the number is 64 = 71 (b9); in base 12, the number is 110 = 92 (b12). In base 2, when n = 178, n1 = n2 = 6181 and when n = 179, n1 = n2 = 267684506.
If a(81) exists, it is larger than 5*10^14. - Giovanni Resta, Jul 21 2017

			n1(1) = 12, n2(1) = 0
n1(2) = 12 = 12 + digsum(0), n2(2) = 3 = 0 + digsum(12)
n1(3) = 15 = 12 + digsum(3), n2(3) = 6 = 3 + digsum(12)
n1(4) = 21 = 15 + digsum(6), n2(4) = 12 = 6 + digsum(15)
n1(5) = 24 = 21 + 3, n2(5) = 15 = 12 + 3
n1(6) = 30 = 24 + 6, n2(6) = 21 = 15 + 6
n1(7) = 33 = 30 + 3, n2(7) = 24 = 21 + 3
n1(8) = 39 = 33 + 6, n2(8) = 30 = 24 + 6
n1(9) = 42 = 39 + 3, n2(9) = 42 = 30 + 12

Anthony Sand, Table of n, a(n) for n = 1..80

Cf. A004207.

Table[NestWhileList[{#1 + Total@ IntegerDigits[#2], #2 + Total@ IntegerDigits[#1]} & @@ # &, {n, 0}, UnsameQ @@ # &, 1, 10^4][[-1, -1]], {n, 80}] (* Michael De Vlieger, Jul 17 2017 *)

n1(1) = n, n2(1) = 0, then n1(i) = n1(i-1) + digsum(n2(i-1)), n2(i) = n2(i-1) + digsum(n1(i-1)) until n1(i) = n2(i).

A329068 a(1) = 1; thereafter if the sum of digits of all previous terms up to a(n) is even then a(n+1) = (sum of digits of all previous terms)/2, otherwise a(n+1) = (sum of digits of all previous terms)*3 + 1.

Original entry on oeis.org

1, 4, 16, 6, 9, 82, 112, 124, 24, 27, 190, 220, 232, 42, 45, 298, 59, 66, 72, 460, 490, 88, 96, 622, 652, 115, 712, 742, 130, 132, 135, 838, 149, 156, 162, 1000, 167, 174, 180, 1108, 1138, 196, 204, 207, 1270, 1300, 1312, 222, 225, 1378, 239, 246, 252, 1540, 1570, 268, 276

Offset: 1

Bence Bernáth, Nov 03 2019

Bence Bernáth, Table of n, a(n) for n = 1..100001

Cf. A105801, A004207.

clear all;
length_seq=10000;
sequence(1)=1;
seq_for_digits(1)=sequence(1);
for i1=1:1:length_seq
   if  0==mod(sum(seq_for_digits),2)
        sequence(i1+1)=sum(seq_for_digits)/2;
   else
     sequence(i1+1)=sum(seq_for_digits)*3+1;
   end
     append=num2str(sequence(i1+1))-'0';
     seq_for_digits=[seq_for_digits append];
end
result=transpose(sequence);

lista(nn) = {va = vector(nn); va[1] = 1; sd = sumdigits(va[1]); for (n=2, nn, if (sd % 2, va[n] = 3*sd+1, va[n] = sd/2); sd += sumdigits(va[n]);); va;} \\ Michel Marcus, Nov 04 2019

from itertools import islice
def agen(): # generator of terms
    sd, an = 0, 1
    while True:
        yield an
        sd += sum(map(int, str(an)))
        an = 3*sd+1 if sd&1 else sd//2
print(list(islice(agen(), 60))) # Michael S. Branicky, Nov 12 2022

A341817 If a(n) is odd, add to a(n) its odd digits and iterate; if a(n) is even, add to a(n) its even digits and iterate; if an iteration reproduces a term already in the sequence, don't do this iteration and extend the sequence with the smallest integer not yet present in the sequence.

Original entry on oeis.org

1, 2, 4, 8, 16, 22, 26, 34, 38, 46, 56, 62, 70, 3, 6, 12, 14, 18, 5, 10, 7, 9, 11, 13, 17, 25, 30, 15, 21, 19, 29, 20, 23, 24, 27, 28, 31, 35, 43, 32, 33, 39, 51, 57, 69, 78, 86, 100, 36, 42, 48, 60, 66, 37, 47, 54, 58, 40, 44, 52, 41, 45, 50, 49, 53, 61, 55, 65, 59, 73, 83, 63, 64, 74, 67, 68, 82, 92

Offset: 1

Eric Angelini and Carole Dubois, Feb 20 2021

This sequence is, by definition, a permutation of the positive integers.

			a(1) = 1, odd, a(2) is thus 1 + 1 = 2;
a(2) = 2, even, thus a(3) = 2 + 2 = 4;
a(3) = 4, even, thus a(4) = 4 + 4 = 8;
a(4) = 8, even, thus a(5) = 8 + 8 = 16;
a(5) = 16, even, thus a(6) = 16 + 6 = 22;
a(6) = 22, even, thus a(7) = 22 + 2 + 2 = 26;
...
a(13) = 70, even, thus a(14) = 70 + 0 = 70 (already in the sequence, thus a(14) = 3 instead, the smallest integer not yet present in the sequence);
a(14) = 3, odd, thus a(15) = 3 + 3 = 6;
a(15)  = 6, even, thus a(16) = 6 + 6 = 12;
a(16) = 12, even, thus a(17) =  12 + 2 = 14;
a(17) = 14, even, thus a(18) = 14 + 4 = 18;
a(18) =  18, even, thus a(19) = 18 + 6 = 26 (already in the sequence, thus a(19) = 5 instead, the smallest integer not yet present in the sequence);
a(19) = 5, odd, thus a(20) = 5 + 5 = 10; etc.

Carole Dubois, Table of n, a(n) for n = 1..5000

Cf. A004207 (sum of digits of all previous terms).

A036230 a(n+1) = a(n) + sum of digits of a(n) starting with 110.

Original entry on oeis.org

110, 112, 116, 124, 131, 136, 146, 157, 170, 178, 194, 208, 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, 607, 620, 628, 644, 658, 677, 697, 719

Offset: 1

Miklos SZABO (mike(AT)ludens.elte.hu)

Elements >= 218 can be found in A004207.

Cf. A004207, A016052, A007618, A006507, A016052.

NestList[#+Total[IntegerDigits[#]]&,110,50] (* Harvey P. Dale, Oct 01 2017 *)

A036231 a(n+1) = a(n) + sum of digits of a(n) starting with 121.

Original entry on oeis.org

121, 125, 133, 140, 145, 155, 166, 179, 196, 212, 217, 227, 238, 251, 259, 275, 289, 308, 319, 332, 340, 347, 361, 371, 382, 395, 412, 419, 433, 443, 454, 467, 484, 500, 505, 515, 526, 539, 556, 572, 586, 605, 616, 629, 646, 662, 676, 695, 715, 728, 745

Offset: 1

Miklos SZABO (mike(AT)ludens.elte.hu)

Elements >= 1003 can be found in A004207.

Cf. A004207, A016052, A007618, A006507, A016052.

A036232 a(n+1) = a(n) + sum of digits of a(n) starting with 211.

Original entry on oeis.org

211, 215, 223, 230, 235, 245, 256, 269, 286, 302, 307, 317, 328, 341, 349, 365, 379, 398, 418, 431, 439, 455, 469, 488, 508, 521, 529, 545, 559, 578, 598, 620, 628, 644, 658, 677, 697, 719, 736, 752, 766, 785, 805, 818, 835, 851, 865, 884, 904, 917, 934

Offset: 1

Miklos SZABO (mike(AT)ludens.elte.hu)

Elements >= 317 can be found in A007618.

Cf. A004207, A016052, A007618, A006507, A016052.

NestList[#+Total[IntegerDigits[#]]&,211,50] (* Harvey P. Dale, Jul 16 2020 *)

A112436 Next term is the sum of the last 10 digits in the sequence, beginning with a(10) = 5.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 5, 10, 11, 13, 17, 25, 22, 25, 30, 29, 32, 30, 29, 33, 36, 34, 36, 42, 37, 41, 37, 40, 35, 37, 37, 42, 38, 45, 46, 46, 46, 50, 44, 43, 40, 34, 31, 30, 25, 25, 28, 31, 31, 32, 30, 26, 24, 26, 30, 28, 35, 35, 37, 39, 48, 50, 47, 50, 45, 42, 36, 40, 33

Offset: 1

Alexandre Wajnberg, Dec 11 2005

Variation on Angelini's A112395. The sequence cycles at a(28)=42 and the loop has 312 terms. Computed by Gilles Sadowski.

			a(28)=42 because 2+9 + 3+3 + 3+6 + 3+4 + 3+6 = 42

Cf. A112395, A004207, A065075, A001370.

cp's OEIS Frontend

A248078 a(1) = 1; a(n+1) = a(n) + product of digits of a(n) + sum of digits of a(n).

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A272233 Number of steps before n1(i) = n2(i) when n1(i) = n1(i-1) + digsum(n2(i-1)), n2(i) = n2(i-1) + digsum(n1(i-1)) and n1(1) = 10^(n-1), n2(1) = 0.

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A272235 In base 2, number of steps before n1(i) = n2(i) when n1(i) = n1(i-1) + digsum(n2(i-1)), n2(i) = n2(i-1) + digsum(n1(i-1)) and n1(1) = 2^(n-1), n2(1) = 0.

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A289979 Define two sequences n1(i) and n2(i) by the recurrences n1(i) = n1(i-1) + digsum(n2(i-1)), n2(i) = n2(i-1) + digsum(n1(i-1)), with initial values n1(1) = n and n2(1) = 0. Then a(n) is the smallest m such that n1(i) = n2(i) = m for some i, or -1 if no such m exists.

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A329068 a(1) = 1; thereafter if the sum of digits of all previous terms up to a(n) is even then a(n+1) = (sum of digits of all previous terms)/2, otherwise a(n+1) = (sum of digits of all previous terms)*3 + 1.

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A341817 If a(n) is odd, add to a(n) its odd digits and iterate; if a(n) is even, add to a(n) its even digits and iterate; if an iteration reproduces a term already in the sequence, don't do this iteration and extend the sequence with the smallest integer not yet present in the sequence.

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A036230 a(n+1) = a(n) + sum of digits of a(n) starting with 110.

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A036231 a(n+1) = a(n) + sum of digits of a(n) starting with 121.

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A036232 a(n+1) = a(n) + sum of digits of a(n) starting with 211.

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