Stuart Coe - cp's OEIS frontend

A378513 a(1) = 1, a(n) = a(n-1) + n if all the digits of a(n-1) do not share a divisor greater than 1. Otherwise, a(n) = a(n-1) divided by the gcd of all its individual digits.

1, 3, 1, 5, 1, 7, 1, 9, 1, 11, 22, 11, 24, 12, 27, 43, 60, 10, 29, 49, 70, 10, 33, 11, 36, 12, 39, 13, 42, 21, 52, 84, 21, 55, 11, 47, 84, 21, 60, 10, 51, 93, 31, 75, 120, 166, 213, 261, 310, 360, 120, 172, 225, 279, 334, 390, 130, 188, 247, 307, 368, 430

Offset: 1

Stuart Coe, Nov 29 2024

It can be shown that this sequence will grow indefinitely: Once it has reached a given number of digits in length, it cannot drop below that number of digits. Regardless of the number of times the number is reduced by dividing by GCDs, n will inevitably be great enough to increase a(n) to a larger and larger number of digits in length.

			a(37) = 47 + 37, because the digits of 47 do not share a factor greater than 1.
a(38) = 84 / 4 = 21, because the gcd of 8 and 4 is 4.

Paolo Xausa, Table of n, a(n) for n = 1..10000

Cf. A052423 (gcd of digits).

a:= proc(n) option remember; `if`(n=1, 1, (t-> (g->
      `if`(g=1, t+n, t/g))(igcd(convert(t, base, 10)[])))(a(n-1)))
    end:
seq(a(n), n=1..62);  # Alois P. Heinz, Nov 29 2024

Module[{n = 1, g}, NestList[If[n++; (g = GCD @@ IntegerDigits[#]) == 1, # + n, #/g] &, 1, 100]] (* Paolo Xausa, Dec 16 2024 *)

lista(nn) = my(v=vector(nn)); v[1] = 1; for (n=2, nn, my(g=gcd(digits(v[n-1]))); if (g == 1, v[n] = v[n-1]+n, v[n] = v[n-1]/g);); v; \\ Michel Marcus, Dec 09 2024

A378505 a(0) = 0, a(n) = 0 where a(n-1) is nonzero and divisible by n. Otherwise a(n) = n + a(n-1).

Original entry on oeis.org

0, 1, 3, 0, 4, 9, 15, 22, 30, 39, 49, 60, 0, 13, 27, 42, 58, 75, 93, 112, 132, 153, 175, 198, 222, 247, 273, 300, 328, 357, 387, 418, 450, 483, 517, 552, 588, 625, 663, 0, 40, 81, 123, 166, 210, 255, 301, 348, 396, 445, 495, 546, 598, 651, 705, 760, 816

Offset: 0

Stuart Coe, Nov 29 2024

From Robert Israel, Nov 29 2024: (Start)
There are an infinite number of zeros in the sequence. If not, suppose a(m) was the last one. Then for j > m, we would have a(j-1) = (m+1) + (m+2) + ... + (j-1) = (j+m)*(j-m-1)/2. In particular, a(m*(m+1)-1) = (m-1)*m*(m+1)*(m+2)/2, which is divisible by m*(m+1) since m-1 or m+2 is even, and that would mean a(m*(m+1)) = 0, contradiction. (End)
From David A. Corneth, Nov 30 2024: (Start)
For some z_0 > 0 such that a(z_0) = 0 we can find the next z_1 > z_0 such that a(z_1) = 0 but for any z_0 < t < z_1 we have a(t) > 0. Then a(t) = binomial(t+1, 2) - binomial(z0+1, 2). Since a(z_1) = 0 we have z_1 | (binomial(z_1 -1 +1, 2) - binomial(z0+1, 2)) = z_1 * (z_1 - 1) / 2 - z_0 * (z_0 + 1)/2.
This means z_1 | 2*(z_1 * (z_1 - 1) / 2 - z_0 * (z_0 + 1)/2) = (z_1 * (z_1 - 1) - z_0 * (z_0 + 1)) where (z_1 * (z_1 - 1) - z_0 * (z_0 + 1)) > 0. As z_1 | z_1 * (z_1 - 1) we also have z_1 | (z_0 * (z_0 + 1)). By the proof of Robert Israel above such z_1 must exist. (End)

			a(11) = 49 + 11 = 60, since a(10) = 49, which does not exactly divide by 11.
a(12) = 0, since a(11) = 60, a nonzero number that exactly divides by 12.
a(13) = 0 + 13 = 13, since a(12) = 0 (so is not nonzero).
From _David A. Corneth_, Nov 30 2024: (Start)
The next 0 after n = 12 is a divisor of 12 * (12 + 1) = 156. The divisors of 156 that are larger than 12 + 1 = 13 are 26, 39, 52, 78, 156. Of these, 39 is the smallest z_1 such that z_1 | z_1 * (z_1 - 1) / 2 - z_0 * (z_0 + 1)/2. Therefore the next 0 after n = 12 is at n = 39.
The first few zeros are at 0, 3, 12, 39, 65, 78, 158, 237, ... (= A379013). As 100 is between the consecutive zeros 78 and 158 there is no zero strictly between 78 and 158 and so a(100) = 100*(100+1) / 2 - 78*(78+1)/2 = 1969. (End)

Paolo Xausa, Table of n, a(n) for n = 0..10000
David A. Corneth, PARI programs
Michael De Vlieger, Log log scatterplot of a(n), n = 1..2^20, showing a(n) = 0 instead as 1/2 for visibility.

Cf. A068627, A379013 (positions of zeros).

a:= proc(n) option remember; `if`(n=0, 0, (t->
     `if`(t>0 and irem(t,n)=0, 0, t+n))(a(n-1)))
    end:
seq(a(n), n=0..56);  # Alois P. Heinz, Nov 29 2024

Module[{n = 0}, NestList[If[Divisible[#, ++n] && # > 0, 0, # + n] &, 0, 100]] (* Paolo Xausa, Dec 09 2024 *)

PARI
```
\\ See Corneth link
```

from itertools import count, islice
def agen(): # generator of terms
    an = 0
    for n in count(1):
        yield an
        an = 0 if an > 0 and an%n == 0 else an + n
print(list(islice(agen(), 70))) # Michael S. Branicky, Nov 29 2024

A376930 a(0)=0, a(1)=1; for n>1, a(n) = a(n-1)+a(n-2), except where a(n-1) is a prime greater than 2, in which case a(n) = a(n-1)-a(n-2).

Original entry on oeis.org

0, 1, 1, 2, 3, 1, 4, 5, 1, 6, 7, 1, 8, 9, 17, 8, 25, 33, 58, 91, 149, 58, 207, 265, 472, 737, 1209, 1946, 3155, 5101, 1946, 7047, 8993, 16040, 25033, 8993, 34026, 43019, 8993, 52012, 61005, 113017, 52012, 165029, 217041, 382070, 599111, 981181, 1580292, 2561473

Offset: 0

Stuart Coe, Oct 11 2024

nonn

It is not clear whether this sequence continues to grow or whether it become stuck in a loop (which could happen if two primes occur in terms n and n-1 or terms n and n-2). Indeed, the sequence is stuck in a loop from around n=10 if we do not ignore the prime number 2.
Similarly, it is not known if the sequence contains any negative terms (which may happen if two primes are adjacent or separated by one other term).
If it continues to grow, it is not clear whether this sequence will contain an infinite number of prime numbers.
Beyond the trivial case of 1, it is not clear if any number will appear more than three times in the sequence. 8993 appears three times, due to several prime terms in close succession.
Also 4618239875200356592 appears three times, as a(111), a(114) and a(117). - Robert Israel, Nov 12 2024

			a(2) = a(1) + a(0) [as a(1) is not a prime > 2] = 1 + 0 = 1.
a(3) = a(2) + a(1) [as a(2) is not a prime > 2] = 1 + 1 = 2.
a(4) = a(3) + a(2) [as a(3) is not a prime > 2] = 2 + 1 = 3.
a(5) = a(4) - a(3) [as a(4) is a prime > 2]     = 3 - 2 = 1.

Robert Israel, Table of n, a(n) for n = 0..4810

Cf. A092942, A229137,

f:= proc(n) option remember;
      if procname(n-1) > 2 and isprime(procname(n-1)) then procname(n-1) - procname(n-2)
      else procname(n-1) + procname(n-2)
      fi
end proc:
f(0):= 0: f(1):= 1:
seq(f(i),i=0..100); # Robert Israel, Nov 12 2024

s={0,1};Do[If[PrimeQ[s[[-1]]]&&s[[-1]]>2,AppendTo[s,s[[-1]]-s[[-2]]],AppendTo[s,s[[-1]]+s[[-2]]]  ],{n,48}];s (* James C. McMahon, Nov 07 2024 *)

from sympy import isprime
from itertools import islice
def agen(): # generator of terms
    a = [0, 1]
    yield from a
    while True:
        an = a[-1]+a[-2] if a[-1] < 3 or not isprime(a[-1]) else a[-1]-a[-2]
        yield an
        a = [a[-1], an]
print(list(islice(agen(), 50))) # Michael S. Branicky, Oct 11 2024

A376257 a(n) = (n - c(n))*(-1)^c(n), where c(n) is the product of the digits of n (A007954).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 10, -10, 10, -10, 10, -10, 10, -10, 10, -10, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 30, -28, 26, -24, 22, -20, 18, -16, 14, -12, 40, 37, 34, 31, 28, 25, 22, 19, 16, 13, 50, -46, 42, -38, 34, -30, 26, -22, 18, -14, 60, 55, 50, 45, 40, 35, 30, 25, 20, 15, 70, -64, 58

Offset: 1

Stuart Coe, Sep 17 2024

There is an interesting pattern on the graph of the sequence.

			For n = 129: 1*2*9=18. So a(n) = (129-18)*(-1)^18 = 111.

Paolo Xausa, Table of n, a(n) for n = 1..10000

Cf. A007954, A070565.

A376257[n_] := (n - #)*(-1)^# & [Times @@ IntegerDigits[n]];
Array[A376257, 100] (* Paolo Xausa, Dec 10 2024 *)

a(n) = (n-A007954(n)) * (-1)^A007954(n).

A376129 Run lengths of the most significant decimal digit in the primes (A077648).

Original entry on oeis.org

1, 1, 1, 1, 4, 2, 2, 3, 2, 2, 3, 2, 1, 21, 16, 16, 17, 14, 16, 14, 15, 14, 135, 127, 120, 119, 114, 117, 107, 110, 112, 1033, 983, 958, 930, 924, 878, 902, 876, 879, 8392, 8013, 7863, 7678, 7560, 7445, 7408, 7323, 7224, 70435, 67883, 66330, 65367, 64336, 63799, 63129, 62712, 62090

Offset: 1

Stuart Coe, Sep 11 2024

			The primes and the run lengths of their initial digits begin
  primes   2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, ...
  runs                 \------------/  \----/  \----/
  lengths  1  1  1  1         4          2       2     ...

Cf. A000720, A037124, A077648.

from sympy import primepi
def A376129(n):
    if n<5: return 1
    def f(m): return (lambda x:primepi(10**x[0]*(x[1]+1)))(divmod(m,9))
    return int(f(n+5)-f(n+4)) # Chai Wah Wu, Oct 16 2024

a(n) = A000720(A037124(n+6)) - A000720(A037124(n+5)) for n >= 5. - Pontus von Brömssen, Oct 07 2024

A376252 Concatenated (n+1)||n modulo n.

Original entry on oeis.org

0, 0, 1, 2, 0, 4, 3, 2, 1, 0, 1, 4, 9, 2, 10, 4, 15, 10, 5, 0, 16, 12, 8, 4, 0, 22, 19, 16, 13, 10, 7, 4, 1, 32, 30, 28, 26, 24, 22, 20, 18, 16, 14, 12, 10, 8, 6, 4, 2, 0, 49, 48, 47, 46, 45, 44, 43, 42, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20

Offset: 1

Stuart Coe, Sep 17 2024

There is an interesting and striking pattern in the graph of this sequence that appears at n >= 20 and appears to continue indefinitely.
There does not appear to be a corresponding pattern for other bases.
Beyond the first two terms, zeros only appear where n is a multiple of 5.

			For n=2: 32 mod 2 is 0.
For n=123: 124123 mod 123 is 16.

Cf. A055642, A127423.

a[n_]:=Mod[FromDigits[Join[IntegerDigits[n+1],IntegerDigits[n]]],n]; Array[a,80] (* Stefano Spezia, Sep 18 2024 *)

a(n) = eval(concat(Str(n+1), Str(n))) % n; \\ Michel Marcus, Sep 17 2024

a(n) = 10*10^logint(n, 10) % n; \\ Ruud H.G. van Tol, Oct 26 2024

def a(n): return int(str(n+1)+str(n))%n
print([a(n) for n in range(1, 81)]) # Michael S. Branicky, Sep 17 2024

def A376252(n): return int('1'+str(n))%n # Chai Wah Wu, Oct 01 2024

a(n) = 10^A055642(n) mod n. Concatenation of 1||n modulo n. - Chai Wah Wu, Oct 01 2024

cp's OEIS Frontend

User: Stuart Coe

A378513 a(1) = 1, a(n) = a(n-1) + n if all the digits of a(n-1) do not share a divisor greater than 1. Otherwise, a(n) = a(n-1) divided by the gcd of all its individual digits.

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A378505 a(0) = 0, a(n) = 0 where a(n-1) is nonzero and divisible by n. Otherwise a(n) = n + a(n-1).

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A376930 a(0)=0, a(1)=1; for n>1, a(n) = a(n-1)+a(n-2), except where a(n-1) is a prime greater than 2, in which case a(n) = a(n-1)-a(n-2).

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A376257 a(n) = (n - c(n))*(-1)^c(n), where c(n) is the product of the digits of n (A007954).

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A376129 Run lengths of the most significant decimal digit in the primes (A077648).

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Python

Formula

A376252 Concatenated (n+1)||n modulo n.

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Mathematica

PARI

PARI

Python

Python

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