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A347475
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Numbers k such that k and the k-th triangular number T(k) = k*(k+1)/2 have only odd digits.
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4
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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
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OFFSET
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1,2
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COMMENTS
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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
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LINKS
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FORMULA
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EXAMPLE
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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.
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MATHEMATICA
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q[n_] := AllTrue[IntegerDigits[n], OddQ]; Select[Range[10^6], And @@ q /@ {#, #*(# + 1)/2} &] (* Amiram Eldar, Nov 20 2021 *)
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PROG
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(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.
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)))
(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)
"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
N=1 # Example of use of the above function:
for n in range(30): print(N := next_A347475(N), end=", ")
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CROSSREFS
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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).
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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