A347475 Numbers k such that k and the k-th triangular number T(k) = k*(k+1)/2 have only odd digits.

1, 5, 13, 17, 177, 1777, 3937, 5537, 5573, 15173, 55377, 55733, 79137, 135173, 195937, 339173, 377777, 399377, 791377, 3397973, 5199137, 7913777, 13535137, 17397537, 33993973, 37735377, 39993777, 59591173, 59919137, 79971937, 135157537, 139713973, 153177777

Offset: 1

Author

M. F. Hasler, Nov 20 2021

Keywords

• nonn
• base

Comments

There is only 1 term with 3 digits and there are only 3 terms with 7 digits. It appears that this (7 digits) is the only length where no term starts with digit 1, and for any length L > 9, the smallest L-digit term (cf. A349247) starts with digits "119...".

Can it be proved that the number of L-digit terms (cf. A355276) tends to infinity as L -> oo?

Can it be proved (or disproved) that the sequence of initial digits of the smallest L-digit term A349247(L) converge, maybe to (1, 1, 9, 3, 1, 1, ...)?

The sequence contains all numbers of the form 33(9{n}7){k}3{n}, where {x} means to repeat the preceding digit or parenthesized sequence of digits x times, for n >= 1 and k = 2, 3 or 4, and for k = 5 with only one initial '3'. - M. F. Hasler, Sep 10 2022

The sequence also contains the infinite subsequence s(k) = 4*10^(1+2*k) - 10^(1+k) - 10^(2+2*k) + 34*10^(3+3*k) + (22*10^k-1)/3. - Kebbaj Mohamed Reda, Sep 11 2022

In the notation of the earlier comment, the above s(k) = 339{k+1}39{k}73{k}. - M. F. Hasler, Sep 13 2022

Examples

			The numbers k = 1, 5, 13, 17, 177, 1777, ... have only odd digits, and the associated triangular numbers T(k) = k*(k+1)/2 = 1, 15, 91, 153, 15753, 1579753, 7751953, ... also have only odd digits.
The same is true for k = 119311115937719393371311137, the smallest 27-digit term.
Any number of the form n = 339{k}79{k}73{k} yields T(n) = A000217(n) = 79{k}19{k}13{k-1}453{k+1}5{k}1{k} and therefore is in the sequence, where {k} means k times (the preceding digit), for any k >= 1.
		

Links

• M. F. Hasler, Table of n, a(n) for n = 1..500, Sep 08 2022.
• S. S. Gupta, Can You Find (CYF) no. 55, Nov 11 2021, updated Sep 12 2022
• A. Zimmermann, Al Zimmermann's Programming Contests: Oddly Triangular, Sep. 7-8, 2022

Crossrefs

Cf. A000217 (triangular numbers), A014261 (numbers with only odd digits), A117960 (triangular numbers with only odd digits), A349243 (indices of the former), A349247 (least k-digit term), A355277 (largest k-digit term), A355276 (number of k-digit terms).

Programs

Mathematica

q[n_] := AllTrue[IntegerDigits[n], OddQ]; Select[Range[10^6], And @@ q /@ {#, #*(# + 1)/2} &] (* Amiram Eldar, Nov 20 2021 *)

PARI

apply( {A347475_row(n, t=10^n\9, L=List())=forvec(v=vector(n,i,[0,4]), is_A014261((1+n=t+fromdigits(v)*2)*n\2)&& listput(L,n));L}, [1..8]) \\ row(n) = terms with n digits. Use concat(%) to flatten the list.

PARI

A347475_first(n)=vector(n,i, n = next_A347475(n*(i>1)+1))
A347475_next(n)={my(t, p, f(v)=for(i=1, #v, bittest(v[i], 0) || return(10^(#v-i)))); while(((p=f(digits(n))) && !n+=p*10\9+if(p>99,22)-n%p) || p=f(digits(t=n*(n+1)\2)), n=max(sqrtint((t+p*10\9-t%p)*2), n+2));n} \\ used in A349247
A347475_prec(n)={my(t, p, f(v)=for(i=1, #v, bittest(v[i], 0) || return(10^(#v-i)))); while(((p=f(digits(n))) && !n-=n%p+if(p>99 && n\p%10, 23, 3)) || p=f(digits(t=n*(n+1)\2)), n=min(sqrtint((t-t%p-1)*2), n-2); if(n>p=n%100, n+=select(t->t<=p,[77,73,37,33,-23])[1]-p)); n} \\ used in A355277. - M. F. Hasler, Sep 13 2022

Python

from itertools import islice, count, product
def A347345gen(): return filter(lambda k: set(str(k*(k+1)//2)) <= {'1','3','5','7','9'}, (int(''.join(d)) for l in count(1) for d in product('13579',repeat=l)))
A347345_list = list(islice(A347345gen(),30)) # Chai Wah Wu, Dec 05 2021

Python

from math import isqrt
def first_even(n):
    "Return 10^k corresponding to first even digit in n."
    for i,c in enumerate(n := str(n), 1):
        if c in "02468": return 10**(len(n)-i)
def next_A347475(n):
    "Return the least term > n."
    if f := first_even(n := n+1): # next larger having only odd digits
        n += f*10//9 - n % f
    while f := first_even(t := n*(n+1)//2):
        if f := first_even(n := max(isqrt((t + 10*f//9 - t % f)*2), n+2)):
            n += 10*f//9 - n % f
    return n  #  M. F. Hasler, Sep 08 2022
N=1 # Example of use of the above function:
for n in range(30): print(N := next_A347475(N), end=", ")

Formula

Intersection of A014261 and A349243.